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Steven H. Weintraub

Galois Theory

Second Edition

123

Steven H. Weintraub Department of Mathematics Lehigh University Bethlehem, PA, USA

ISBN: 978-0-387-87574-3 e-ISBN: 978-0-387-87575-0DOI: 10.1007/978-0-387-87575-0 Library of Congress Control Number: Mathematics Subject Classification (2000): 12-01, 12F10, 11R32

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002 8937989

Editorial board: Sheldon Axler, San Francisco State University Vincenzo Capasso, Università degli Studi di Milano Carles Casacuberta, Universitat de Barcelona Angus MacIntyre, Queen Mary, University of London Kenneth Ribet, University of California, Berkeley Claude Sabbah, CNRS, École Polytechnique Endre Süli, University of Oxford Wojbor Woyczy ski, Case Western Reserve University

To Judy, after 17 years, and to Blake

Contents

Preface to the First Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix

Preface to the Second Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii

1 Introduction to Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Some Introductory Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Field Theory and Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1 Generalities on Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Extension Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Algebraic Elements and Algebraic Extensions . . . . . . . . . . . . . . 182.5 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.6 Extending Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.7 Normal, Separable, and Galois Extensions . . . . . . . . . . . . . . . . . 252.8 The Fundamental Theorem of Galois Theory . . . . . . . . . . . . . . . 292.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Development and Applications of Galois Theory . . . . . . . . . . . . . . 453.1 Symmetric Functions and the Symmetric Group . . . . . . . . . . . . 453.2 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.4 Disjoint Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.5 Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663.6 The Normal Basis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.7 Abelian Extensions and Kummer Fields . . . . . . . . . . . . . . . . . . . 73

viii Contents

3.8 The Norm and Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 793.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4 Extensions of the Field of Rational Numbers . . . . . . . . . . . . . . . . . 894.1 Polynomials in Q[X ] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 894.2 Cyclotomic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 934.3 Solvable Extensions and Solvable Groups . . . . . . . . . . . . . . . . . . 974.4 Geometric Constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1014.5 Quadratic Extensions of Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1074.6 Radical Polynomials and Related Topics . . . . . . . . . . . . . . . . . . . 1124.7 Galois Groups of Extensions of Q . . . . . . . . . . . . . . . . . . . . . . . . 1224.8 The Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1284.9 Practical Computation of Galois Groups . . . . . . . . . . . . . . . . . . . 1314.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5 Further Topics in Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1435.1 Separable and Inseparable Extensions . . . . . . . . . . . . . . . . . . . . . 1435.2 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1515.3 The Algebraic Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.4 Infinite Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1605.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

6 Transcendental Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1736.1 General Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1736.2 Simple Transcendental Extensions . . . . . . . . . . . . . . . . . . . . . . . . 1816.3 Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1856.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

A Some Results from Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 195A.1 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195A.2 p-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199A.3 Symmetric and Alternating Groups . . . . . . . . . . . . . . . . . . . . . . . 200

B A Lemma on Constructing Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 205

C A Lemma from Elementary Number Theory . . . . . . . . . . . . . . . . . 207

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

Preface to the First Edition

This is a textbook on Galois theory. Galois theory has a well-deserved repu-tation as one of the most beautiful subjects in mathematics. I was seduced byits beauty into writing this book. I hope you will be seduced by its beauty inreading it.

This book begins at the beginning. Indeed (and perhaps a little unusuallyfor a mathematics text), it begins with an informal introductory chapter, Chap-ter 1. In this chapter we give a number of examples in Galois theory, evenbefore our terms have been properly defined. (Needless to say, even thoughwe proceed informally here, everything we say is absolutely correct.) Theseexamples are sort of an airport beacon, shining a clear light at our destinationas we navigate a course through the mathematical skies to get there.

Then we start with our proper development of the subject, in Chapter 2.We assume no prior knowledge of field theory on the part of the reader. Wedevelop field theory, with our goal being the Fundamental Theorem of GaloisTheory (the FTGT). On the way, we consider extension fields, and deal withthe notions of normal, separable, and Galois extensions. Then, in the penulti-mate section of this chapter, we reach our main goal, the FTGT.

Roughly speaking, the content of the FTGT is as follows: To every Galoisextension E of a field F we can associate its Galois group G = Gal(E/F). Bydefinition, G is the group of automorphisms of E that are the identity on F.Then the FTGT establishes a one-to-one correspondence between fields B thatare intermediate between E and F, i.e., between fields B with F ⊆ B ⊆ E, andsubgroups of G. This connection allows us to use the techniques of group the-ory to answer questions about fields that would otherwise be intractable. (In-deed, historically Galois theory has been used to solve questions about fieldsthat were outstanding for centuries, and even for millenia. We will treat someof these questions in this book.) In the final section of this chapter, we returnto the informal examples with which we started the book, as well as treat-

x Preface to the First Edition

ing more intricate and advanced ones that we can handle with our new-foundknowledge.

In Chapter 3 we further develop and apply Galois theory. In this chapterwe deal with a variety of different topics, some of which we mention here.In the first section we use Galois theory to investigate the field of symmet-ric functions and the ring of symmetric polynomials. Galois theory allows usto completely determine the structure of finite fields, and we do this in thethird section. Two important properties of fields are the existence of primitiveelements and normal bases, and we prove these in Sections 3.5 and 3.6. Wecan also say quite a bit about abelian extensions (i.e., extensions with abelianGalois groups), and we treat these in Section 3.7.

We develop Galois theory in complete generality, with careful considera-tion to the situation in positive characteristic as well as in characteristic 0. Butwe are especially interested in algebraic number fields, i.e., finite extensions ofthe field of rational numbers Q. We devote Chapter 4 to considering extensionsof Q. Again we deal with a variety of different topics. We consider cyclotomicpolynomials. We consider the question as to when equations are solvable byradicals (and prove Abel’s theorem that the general equation of degree at least5 is not). We show that the three classical geometric problems of Greek an-tiquity: trisecting the angle, duplicating the cube, and squaring the circle, areunsolvable with straightedge and compass. We deal with quadratic fields andtheir relation to cyclotomic fields, and we deal with radical polynomials, i.e.,polynomials of the form Xn − a, which have a particular theory.

In Chapter 5 we consider more advanced topics in Galois theory. In par-ticular, we prove that every field has an algebraic closure, and that the fieldof complex numbers C is algebraically closed, in Section 5.3, and we developGalois theory for infinite algebraic extensions in Section 5.4.

Note that in Chapters 1 through 4, we assume that all field extensions arefinite, except where explicitly stated otherwise. In Chapter 5 we allow exten-sions to be infinite.

There are three appendices. In the first we develop some necessary grouptheory. (We have put this in the appendix for logical reasons. The main textdeals with field theory, and to develop the necessary group theory would leadto digressions in the main line of argument. Thus we collect these facts in anappendix in order to have a clear line of argument.) The second appendix re-visits some material in the text from a more advanced point of view, which wedo not want to presuppose of the reader, while the third appendix presents anelementary but tricky argument that enables us to avoid relying on Dirichlet’stheorem about primes in an elementary progression at one point.

Our approach has been heavily influenced by Artin’s classic 1944 textGalois Theory. Artin’s approach emphasized linear algebra, and our approach

Preface to the First Edition xi

has the same (and perhaps greater) emphasis. We have tried to have minimalprerequisites for this book, but, given this emphasis, the reader should have asound knowledge of linear algebra. Beyond that the reader should know thebasic facts about groups and rings, and especially about polynomial rings. Asa source for this background material we naturally recommend our previousbook, Algebra: An Approach via Module Theory, by William A. Adkins andSteven H. Weintraub, Springer-Verlag Graduate Texts in Mathematics No. 136.We refer to this text as [AW] when we have occasion to cite it. Also, the readershould be familiar with elementary number theory (the material containedin any standard undergraduate course in the subject will more than suffice).Finally, the Krull topology is the key to understanding infinite Galois exten-sions, so in the final section of this book (Section 5.4), and in this sectionalone, the reader must have a good knowledge of point-set topology.

There is, roughly speaking, enough material in this book for a year-longcourse in Galois theory. A one-semester course could consist of Chapters 1and 2 (both of which can be easily covered in a semester) plus additional ma-terial from Chapters 3 and 4 as interest dictates and time permits.

We would like to mention that our previous book, [AW], treated groups,rings, modules, and linear algebra. This book treats field theory, so togetherthese two books cover the topics of a standard one-year graduate algebracourse. Also, given our particular attention to Galois theory over Q, we feelthis book would be especially well-suited to students with an interest in alge-braic number theory.

Our numbering system in this book is fairly standard. Theorem a.b.c refersto Theorem b.c in section b of Chapter a (or Appendix a). We denote the endof proofs by ��, as usual. In case a result is immediate, we simply appendthis symbol to its statement. Theorems, etc., are in italics, so are naturally setoff from the remaining text. Definitions, etc., are in roman, and so are not.To delimit them, we end them with the symbol �. Our notation is also fairlystandard, but we call the reader’s attention to the following conventions: Weuse A ⊆ B to mean that A is a subset of B, while A ⊂ B means that A isa proper subset of B. We denote fields by boldface letters and the integers byZ. We will often be considering the situation where the field E is an extensionof the field F, and in this situation we will use greek letters (α, β, γ, . . . ) todenote elements of E and roman letters (a, b, c, . . . ) to denote elements ofF. The greek letter ω will denote the primitive complex cube root of unityω = (−1 + i

√3)/2. The letter p will always denote a prime. Finally, id will

denote the identity automorphism of whatever object is under consideration.

June 2005 Steven H. Weintraub

Preface to the Second Edition

I am pleased that there has been enough interest in this book to justify a secondedition in such a short time since the first edition appeared.

The main change in this second edition is that I have added a new chapter,Chapter 6, that deals with transcendental extensions, thereby expanding thescope of the book. In addition, I have added smaller amounts of material: thestatement and proof of Newton’s Identities at the end of Section 3.1, materialon linear disjointness at the end of Section 3.4, an expanded treatment of Ex-ample 2.9.7, and an additional corollary at the end of Section 4.7. Also, thereis one consistent change, purely for esthetic reasons: Defined terms are in ital-ics, instead of boldface, as they were previously. Beyond that, I have taken theopportunity to correct all (known) typos and minor errors in the first edition.(To my knowledge, there were no major errors.) It would be nice to hope thatin adding around 25 pages to the text, I have not introduced any new errors;alas, that is probably too optimistic. Well, one can hope, anyway.

I have made it a point to preserve the numbering from the first editionintact. That is, in all cases Theorem a.b.c from the first edition remains Theo-rem a.b.c in the second edition, etc. Thus readers of the first edition should notbe too disadvantaged.

June 2008 Steven H. Weintraub

1

Introduction to Galois Theory

1.1 Some Introductory Examples

In this section we will proceed informally, neither proving our claims nor evencarefully defining our terms. Nevertheless, as you will see in the course ofreading this book, everything we say here is absolutely correct. We proceed inthis way to show in advance what our main goals are, and hence to motivateour development.

Example 1.1.1. Let F = Q, the field of rational numbers, and consider thepolynomial X2−2 ∈ Q[X ]. We let E be the “splitting field” of this polynomial,i.e., the “smallest” field which contains “all” the roots of this polynomial. ThenE = Q(

√2), the field obtained by “adjoining”

√2 to Q. Then Q(

√2) =

{a + b√

2 | a, b ∈ Q}, so as a vector space, dimF E = 2. We ask for allautomorphisms of E that fix F. If σ : E → E is such an automorphism, then itmust take a root of X2 −2 to a root of X2 −2. Thus there are two possibilities:σ1(

√2) = √

2, in which case σ1 : E → E is the identity map, or σ2(√

2) =−√

2, in which case σ2 : E → E is given by σ2(a+b√

2) = a−b√

2. Thus thegroup of automorphisms of E that fix F, known as the Galois group Gal(E/F),is {σ1 = id, σ2}, isomorphic to Z/2Z, of order 2. Note that | Gal(E/F)| =dimF E. �Example 1.1.2. Let F = Q and consider the polynomial (X2 − 2)(X2 − 3) ∈Q[X ]. The splitting field of this polynomial is E = Q(

√2,

√3) and E =

{a + b√

2 + c√

3 + d√

6 | a, b, c, d ∈ Q}, so dimF E = 4. Again we askfor all automorphisms σ of E that fix F. We must have that σ takes a root ofX2 −2 to a root of X2 −2, so σ(

√2) = ±√

2, and that σ takes a root of X3 −3to X2 − 3, so σ(

√3) = ±√

3. Indeed, the values of σ on√

2 and√

3 can be

S.H. Weintraub, Galois Theory, DOI 10.1007/978-0-387-87575-0_1, © Springer Science+Business Media, LLC 2009

2 1 Introduction to Galois Theory

chosen independently, and these determine the values of σ on all elements ofE, so we have the four automorphisms of E given by

σ1(a + b√

2 + c√

3 + d√

6) = a + b√

2 + c√

3 + d√

6,σ2(a + b

√2 + c

√3 + d

√6) = a − b

√2 + c

√3 − d

√6,

σ3(a + b√

2 + c√

3 + d√

6) = a + b√

2 − c√

3 − d√

6,σ4(a + b

√2 + c

√3 + d

√6) = a − b

√2 − c

√3 + d

√6;

thus, G = Gal(E/F) = {σ1 = id, σ2, σ3, σ4} is isomorphic to Z/2Z ⊕ Z/2Z,and | Gal(E/F)| = 4 = dimF E.

Now we ask for all fields B “intermediate” between E and F, i.e., fields Bcontaining F and contained in E. Of course B1 = F and B5 = E are certainlyintermediate fields. Here are three others that easily appear:

B2 = Q(√

2), B3 = Q(√

3), B4 = Q(√

6).

Now let us look at subgroups H of G. It is easy to list all such H . They areH1 = G, H5 = {id} and also

H2 = {id, σ3}, H3 = {id, σ2}, H4 = {id, σ4}.Note we have a one-to-one correspondence between {Hi } and {Bi } given

by

Bi = Fix(Hi )

where Fix(H) is the subfield of E fixed by all elements of the subgroup H .In fact, the Fundamental Theorem of Galois Theory (FTGT) tells us there isa one-to-one correspondence between subgroups H of Gal(E/F) and fields Bintermediate between E and F given by

H ←→ B where B = Fix(H).

It is easy to determine all the subgroups of G, as this is a finite group,so this gives us a way of determining all intermediate fields (a priori, a muchmore different task), and in our case this shows that B1, . . . , B5 are indeed allthe intermediate fields.

Let us make a further observation here, which the FTGT also tells us istrue in general: If B is an intermediate field, then B is an F-vector space and Eis a B-vector space, and

dimF B = [G : H ], dimB E = |H |.

1.1 Some Introductory Examples 3

One more observation: For each intermediate field B, we may considerGal(B/F), the group of automorphisms of B fixing F. If B = F, thenGal(B/F) = Gal(F/F) consists of the identity element alone, while if B = E,then Gal(B/F) = Gal(E/F) is the whole original Galois group. For B = Bi ,i = 2, 3, 4, Gal(B/F) is isomorphic to Z/2Z, of order 2. Thus in all five cases,

| Gal(B/F)| = dimF B. �Example 1.1.3. Let F = Q and consider the polynomial X3 − 2 ∈ Q[X ].We let E be the splitting field of this polynomial. Then 3

√2 ∈ E and, since

E is a field, (3√

2)2 = 3√

4 ∈ E. Now E contains all the roots of x3 − 2, soif ω is a primitive cube root of 1, then ω

3√

2 ∈ E, and since E is a field,ω

3√

2/3√

2 = ω ∈ E. (Also, ω2 ∈ E, but note that ω3 = 1, ω3 − 1 = 0,(ω − 1)(ω2 + ω + 1) = 0 and ω �= 1, so ω2 + ω + 1 = 0 and ω2 = −1 − ω;thus once we add ω to F we already have ω2 as well.) Then E = Q(ω,

3√

2)

and E = {a + b 3√

2 + c 3√

4 + dω + eω 3√

2 + f ω3√

4 | a, b, c, d, e, f ∈ Q} anddimF E = 6.

Now we wish to determine Gal(E/F). Again, any automorphism of E musttake a root of X3 − 2 to another root of X3 − 2, so σ(

3√

2) = 3√

2, ω3√

2, orω2 3

√2. Also, as we just observed, ω is a root of X2 + X +1, whose other root is

ω2, so σ(ω) = ω or ω2. There are three choices for σ(3√

2) and two choices forσ(ω). We may make these choices independently, and these choices determineσ on all of E, so | Gal(E/F)| = 6.

In particular we have the following elements:

ϕ : E → E with ϕ(3√

2) = ω3√

2, ϕ(ω) = ω,

ψ : E → E with ψ(3√

2) = 3√

2, ψ(ω) = ω2.

It is easy to check that ϕ3 = id and ψ2 = id.Let us compute further:

ϕψ(3√

2) = ϕ(ψ(3√

2)) = ϕ(3√

2) = ω3√

2,

ϕψ(ω) = ϕ(ψ(ω)) = ϕ(ω2) = ω2,

while

ψϕ2(3√

2) = ϕ(ψ2(3√

2)) = ψ(ω2 3√

2) = ω3√

2,

ψϕ2(ω) = ψ(ϕ2(ω)) = ψ(ω) = ω2.

Thus we see that ϕψ = ψϕ2, and so

G = Gal(E/F) = 〈ψ, ϕ | ϕ3 = 1, ψ2 = 1, ϕψ = ψϕ−1〉


is isomorphic to the dihedral group of order 6.Now it is easy to list all subgroups H of G; by the FTGT this gives us all

intermediate fields B. The subgroups H are

H1 = G, H2 = {id, ϕ, ϕ2}, H3 = {id, ψ},H4 = {id, ϕψ}, H5 = {id, ϕ2ψ}, H6 = {id}.

Then we can compute that the corresponding subfields Bi = Fix(Hi ) aregiven by

B1 = Q,

B2 = Q(ω) = {a + dω | a, d ∈ Q},B3 = Q(

3√

2) = {a + b3√

2 + c3√

4 | a, b, c ∈ Q},B4 = Q(ω

3√

2) = {a + c3√

4 + eω3√

2 + cω3√

4 | a, c, e ∈ Q},B5 = Q(ω2 3

√2) = {a + b

3√

2 + bω3√

2 + f ω3√

4 | a, b, f ∈ Q},B6 = Q(ω,

3√

2).

Again we see that for each H and corresponding B,

dimF B = [G : H ], dimB E = |H |.Once again, let us compute Gal(B/F) for each intermediate field B. For

B = B1, Gal(B/F) = {id} and for B = B6, Gal(B/F) = G. More inter-estingly, Gal(B2/F) is isomorphic to Z/2Z, with the nontrivial element τ ob-tained as follows: ω is a root of X2 + X + 1, so τ must take ω to the other rootof x2 + x + 1, which is ω2 = −1 − ω. Thus

τ(a + dω) = a + d(−1 − ω) = (a − d) − dω.

(Actually, B2 and τ have an alternate description. By the quadratic formula,ω = (−1+i

√3)/2 and ω2 = ω = (−1−i

√3)/2. Thus B2 = Q(i

√3) = {a′+

b′i√

3 | a′, b′ ∈ Q} and τ(a′ + b′i√

3) = a′ − b′i√

3, so τ is simply complexconjugation on B2). In each of these three cases we see that | Gal(B/F)| =dimF B.

On the other hand, consider B3. An automorphism of B3 = Q(3√

2) fixingQ is determined by its effect on 3

√2. Since 3

√2 is a root of X3 − 2, it must take

3√

2 to a root of this same polynomial. But 3√

2 is the only root of X3 − 2 inB3, so any automorphism of B3 must fix 3

√2 and hence must be the identity.

A similar argument works for B4 and B5, so we see that 1 = | Gal(Bi/F)| <

dimF Bi = 3 for i = 3, 4, 5. Thus the situation here is different.

1.1 Some Introductory Examples 5

Why is it different? The answer is that the subgroups H1, H2, and H6 ofG are normal subgroups of G, while the subgroups H3, H4, and H5 are not.As we will see from the FTGT, in case H is a normal subgroup of G andB = Fix(H), then | Gal(B/F)| = dimF B and, moreover, Gal(B/F) is the quo-tient group G/H . You can verify that this is true here for H1 and H6 (whichis immediate) and for H2 (which is more interesting). Also, we’ll remark thatH3, H4, and H5 are conjugate subgroups of G and that B3, B4 and B5 are “con-jugate” intermediate fields, in that there are automorphisms of E fixing F andpermuting these subfields; indeed ϕ(B3) = B4 and ϕ2(B3) = B5.

Note that in Example 1.1.2 the Galois group G was abelian, so everysubgroup was normal, and hence for every intermediate field B, we had| Gal(B/F)| = dimF B and in fact Gal(B/F) = G/H . �Example 1.1.4. Let F = Q and consider the polynomial X2 − 5 ∈ Q[X ].The splitting field of this polynomial is B = Q(

√5) and the Galois group

Gal(B/F) = {id, ϕ} where ϕ(√

5) = −√5. Now consider the polynomial

5(X) = (X5 − 1)/(X − 1) = X4 + X3 + X2 + X + 1. This polynomial isirreducible. Let E be the splitting field of this polynomial. Then E = Q(ζ5)

where ζ5 = exp(2π i/5), a fifth root of 1, and dimF E = 4. The automorphismof E fixing F must take ζ5 to some other root of 5(X), and any root is pos-sible, as 5(X) is irreducible. Now 5(X) has roots ζ5, ζ 2

5 , ζ 35 , and ζ 4

5 . Letσ ∈ Gal(E/F) with σ(ζ5) = ζ 2

5 . Then σ 2(ζ5) = ζ 45 and σ 3(ζ5) = ζ 8

5 = ζ 35 .

By the FTGT, | Gal(E/F)| = 4 so Gal(E/F) = {id, σ, σ 2, σ 3} is cyclic oforder 4.

Now 0 = 5(ζ5) = ζ 45 + ζ 3

5 + ζ 25 + ζ5 + 1, so (ζ5 + ζ 4

5 )2 = ζ 25 + ζ 3

5 + 2 =ζ5 + ζ 4

5 + 1, so θ = ζ5 + ζ 45 is a root of X2 − X − 1, as is θ ′ = ζ 2

5 + ζ 35 . By

the quadratic formula θ = (1 +√5)/2 and θ ′ = (1 −√

5)/2. Thus we see thatE ⊃ Q(θ) = Q(

√5) = B, and dimB E = 2 as θ = ζ5 + ζ 4

5 = ζ5 + ζ−15 , so

ζ5 is a root of the quadratic equation X2 − θ X + 1, a polynomial in B[X ]. (Itsother root is ζ−1

5 .)By the FTGT, Gal(E/B) has order 2, and is a subgroup of Gal(E/F). Then

it must be the subgroup {id, σ 2}, and that checks as σ(θ) = σ(ζ5 + ζ 45 ) =

ζ 25 + ζ 3

5 = θ ′ so σ 2 does not fix B, but σ 2(θ) = θ (and σ 2(θ ′) = θ ′) so σ 2

does. Also, Gal(B/Q) = Gal(E/Q)/ Gal(E/B) = {id, σ, σ 2, σ 3}/{id, σ 2} hasσ as a representative of the nontrivial element. The quotient map is given byrestricting an automorphism of E to B. We have already seen that Gal(B/Q) ={id, ϕ}. But note that σ | B = ϕ as σ

((1+√

5)/2) = (1−√

5/2), so σ(√

5) =−√

5 = ϕ(√

5).In this example, we are back to the case in which Gal(E/F) is abelian, so

by the FTGT, any intermediate field B will also be a Galois extension of F,i.e., will have | Gal(B/F)| = dimF B. Since Gal(E/F) is cyclic of order 4, it


has a unique subgroup of order 2 corresponding to a unique intermediate fieldB with dimB E = 2, and we have explicitly identified this field B as Q(

√5).

Also, we have shown how to obtain E in two stages: First, adjoin θ (or√

5)

to Q, creating an extension of degree 2, and the adjoin ζ5 to B, creating anextension E of B degree 2, and hence an extension E of Q of degree 4. (Thedegree of an extension is its dimension as a vector space over the base field.)

Finally, this example is interesting as it is an example of the relationshipbetween a quadratic field B (i.e., a field obtained by adjoining a square rootto Q) and a cyclotomic field E (i.e., a field obtained by adjoining a root of1 to Q). �

2

Field Theory and Galois Theory

2.1 Generalities on Fields

We begin by defining the objects we will be studying.

Definition 2.1.1. F is a field if F is an abelian group under addition, F − {0}is an abelian group under multiplication, and multiplication distributes overaddition.

In other words, F is a field if :

1. For any a, b ∈ F, a + b ∈ F.2. For any a, b ∈ F, a + b = b + a.3. For any a, b, c ∈ F, (a + b) + c = a + (b + c).4. There is a 0 ∈ F such that a + 0 = 0 + a = a for every a ∈ F.5. For every a ∈ F there is an element −a ∈ F with a+(−a)=(−a)+a = 0.6. For any a, b ∈ F, ab ∈ F, and if a �= 0 and b �= 0, then ab �= 0.7. For any a, b ∈ F, ab = ba.8. For any a, b, c ∈ F, (ab)c = a(bc).9. There is a 1 ∈ F such that a1 = 1a = a for every a ∈ F.

10. For every a �= 0 ∈ F there is an element a−1 ∈ F with aa−1 = a−1a = 1.11. For every a ∈ F, a0 = 0a = 0.12. For every a, b, c ∈ F, (a + b)c = ac + bc.13. For every a, b, c ∈ F, c(a + b) = ca + ba.14. 0 �= 1. �Example 2.1.2. Here are some examples of fields. The first three are veryfamiliar.


8 2 Field Theory and Galois Theory

(1) The rational numbers Q.(2) The real numbers R.(3) The complex numbers C.(4) The field F = Q(

√D) where D is not a perfect square.

By definition, F = {a +b√

D | a, b ∈ Q} with addition and multiplicationdefined “as usual”. That is, (a + b

√D)+ (c + d

√D) = (a + c)+ (b + d)

√D

and (a + b√

D)(c + d√

D) = (ac + bd D) + (ad + bc)√

D. We need to seethat inverses exist, and this comes from the familiar method of “rationalizingthe denominator”:

1

a + b√

D= 1

a + b√

D· a − b

√D

a − b√

D= a − b

√D

a2 − b2 D

= a

a2 − b2 D− b

a2 − b2 D

√D.

(5) The fields F = Fp of integers modulo p, where p is a prime. Fp ={0, 1, . . . , p−1} with addition and multiplication defined mod p, i.e., i+ j = kin Fp if i + j ≡ k(mod p) and i j = k in Fp if i j ≡ k(mod p). �

Example 2.1.2 (4) illustrates a construction of particular importance to us.It is, as we shall see below, an “algebraic extension” of Q, obtained by “adjoin-ing” the element

√D, an element that is a root of the irreducible polynomial

X2 − D ∈ Q[X ]. (In a similar way, C = R(i) where i is a root of the ir-reducible polynomial X2 + 1 ∈ R[X ]). In fact, this chapter will be entirelydevoted to studying algebraic extensions of fields and their properties.

In this book, we shall not only be studying individual fields, but also, andespecially, maps between them. We thus make the following important obser-vation.

Lemma 2.1.3. Let F and F′ be fields and let ϕ : F → F′ be a map withϕ(a + b) = ϕ(a) + ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b) for every a, b ∈ F. Thenone of the following two alternatives holds:

(1) ϕ(0) = 0, ϕ(1) = 0, and ϕ is the zero map.(2) ϕ(0) = 0, ϕ(1) = 1, and ϕ is an injection.

Proof. We have that ϕ(0) = ϕ(0 + 0) = ϕ(0) + ϕ(0) so 0 = ϕ(0).We also have that ϕ(1) = ϕ(1 · 1) = ϕ(1)ϕ(1) so 0 = ϕ(1)ϕ(1) − ϕ(1) =

ϕ(1)(1−ϕ(1)) and hence, by the contrapositive of property (6) of a field, eitherϕ(1) = 0 or ϕ(1) = 1.

First, suppose ϕ(1) = 0. Then for any a ∈ F, we have that ϕ(a) =ϕ(a · 1) = ϕ(a)ϕ(1) = ϕ(a) · 0 = 0, and ϕ is the zero map.

2.1 Generalities on Fields 9

Next suppose ϕ(1) = 1. Then for any c ∈ F, c �= 0, we have that 1 =ϕ(1) = ϕ(cc−1) = ϕ(c)ϕ(c−1) so ϕ(c) �= 0. Now let a, b ∈ F with a �= b,and set c = b−a �= 0. Then ϕ(b) = ϕ(a+(b−a)) = ϕ(a+c) = ϕ(a)+ϕ(c) �=ϕ(a), and ϕ is an injection. ��Definition 2.1.4. The characteristic char(F) of the field F is the smallest pos-itive integer n such that n · 1 = 0 ∈ F, or 0 if no such n exists. �

Examining Example 2.1.2, we see that the fields in parts (1)–(4) have char-acteristic 0, while the fields in part (5) have characteristic p.

Lemma 2.1.5. Let F be a field. Then char(F) is either 0 or a prime.

Proof. Suppose char(F) = n is neither 0 nor a prime, and let n = ab with1 < a, b < n. Then a ·1 �= 0 and b·1 �= 0 but (a ·1)(b·1) = (ab)·1 = n ·1 = 0.But this is impossible as the product of any two nonzero elements in a field isnonzero. ��Remark 2.1.6. If F is any field, there is a canonical map ϕ from the integers Zto F determined by ϕ(1) = 1. If char(F) = 0, then ϕ is an injection; using ϕ

we regard Z as a subring of F, and then we also regard F0 = Q as a subfieldof F. If char(F) = p, then ϕ gives an injection from Fp into F and we regardF0 = Fp as a subfield of F. In either case, F0 is called the prime field. �Definition 2.1.7. A field E is an extension of the field F if F is a subfieldof E. �

We will often be in the situation of considering a field F and an extensionfield E. To help the reader keep track of what is going on, we will adopt thenotational convention that, in this situation, we will denote elements of F byRoman letters (a, b, . . . ) and elements of E by Greek letters (α, β, . . . ).

It easy to check that if E is an extension of F, then char(E) = char(F) andfurthermore E is an F-vector space. This leads us to the following definition.

Definition 2.1.8. (1) The degree (E/F) of E over F is dimF E, the dimensionof E as an F-vector space.

(2) E is a finite extension of F (or E/F is finite) if (E/F) is finite. �(Although we will not use it here, another common notation for (E/F) is

[E : F].)Examining Example 2.1.2 (4), we see that (Q(

√D)/Q) = 2 as Q(

√D)

has basis {1,√

D} as a Q-vector space (and similarly (C/R) = 2 as C has basis{1, i} as an R-vector space). Note that 2 is also the degree of the polynomial


X2− D (and of the polynomial X2+1). As we shall see, this is no coincidence.On the other hand, (C/Q) = ∞. As we shall also see, with considerably morework, for every prime p and every positive integer n there is a field Fpn withpn elements, and Fpn is unique up to isomorphism. Then (Fpn /Fp) = n.

Here is an important observation about arithmetic in fields of characteris-tic p.

Lemma 2.1.9. Let F be a field of characteristic p. Then (a + b)p = a p + bp

for any elements a, b of F.

Proof. By the Binomial Theorem, (a + b)p = �pi=0

(pi

)ai bp−i = a p + bp as(p

i

) = p!/(i !(p − i)!) is divisible by p for all 1 ≤ i ≤ p − 1. ��Definition 2.1.10. Let F be a field of characteristic p. The Frobenius map : F → F is the map defined by (X) = X p. �Corollary 2.1.11. Let F be a field of characteristic p. Then : F → F is anendomorphism. If F is finite, is an automorphism.

Proof. Clearly (1) = 1 and (ab) = (a)(b). But also (a + b) =(a) + (b) by Lemma 2.1.9. Now, by Lemma 2.1.3, is injective, so| Im()| = |F|. Thus, if |F| is finite, Im() = F. ��Lemma 2.1.12. Let F = Fp. Then : F → F is the identity automorphism.

Proof. By Fermat’s Little Theorem, x p ≡ x(mod p) for x = 0, 1, . . . ,

p − 1. ��Thus, for the field Fp, the Frobenius automorphism is uninteresting. But,

as we shall see, for other fields of characteristic p it plays an important role.We shall find it convenient to introduce the following notion.

Definition 2.1.13. An integral domain is a ring R that satisfies properties (1)–(9) and (11)–(14) of Definition 2.1.1. �

In particular, in an integral domain R, property (6) holds, so the product ofany two nonzero elements of R is nonzero. However, property (11) may nothold, so a nonzero element of R may not have an inverse. Those elements ofR that do have inverses are called the units of R. As an example of an integraldomain, we have the integers Z, from which the name integral domain arises.

2.2 Polynomials 11

2.2 Polynomials

In this section we consider polynomial rings. We shall assume the reader hasalready encountered them, and so will not prove the more basic facts. As areference for them, we recommend [AW, Sections 2.4 and 2.5].

For any field F, we let F[X ] be the ring of polynomials in the variableX with coefficients in F. We observe that the product f (X)g(X) of any twononzero polynomials f (X), g(X) ∈ F[X ] is nonzero, and so F[X ] is an inte-gral domain. We regard F ⊂ F[X ] by identifying the element a ∈ F with theconstant polynomial a ∈ F[X ], and observe that, under this identification, theunits of F[X ] (i.e., the invertible elements of F[X ]) are precisely the nonzeroelements of F.

The most basic fact about F[X ] is that we have the division algorithm: Forany polynomials f (X) and g(X) ∈ F[X ] with g(X) �= 0, there exist uniquepolynomials q(X) and r(X) with

f (X) = g(X)q(X) + r(X),

where either r(X) = 0 or r(X) is a nonzero polynomial with deg(r(X)) <

deg(g(X)). (Here deg is the degree of a polynomial.) We call q(X) the quotientand r(X) the remainder. Note that the quotient and remainder are independentof the field: If E is an extension of F and f (X) = g(X)q(X) + r(X) is anequation in F[X ], this equation holds in E[X ] as well, so by the uniqueness ofthe quotient and remainder, q(X) and r(X) are the quotient and remainder inE[X ] as well.

From the existence of the division algorithm we conclude that F[X ] is aEuclidean domain, where the norm of a nonzero polynomial is equal to itsdegree. A Euclidean domain is so named because it is an integral domain inwhich Euclid’s algorithm holds. One of the consequences of Euclid’s algo-rithm is that any two nonzero elements of F[X ] have a greatest common di-visor (gcd). In general, the gcd of two elements of a Euclidean ring is onlywell defined up to multiplication by a unit, but in F[X ] we specify the gcdby requiring it to be monic. Then the greatest common divisor (gcd) of twopolynomials f (X) and g(X) is defined to be the unique monic polynomialof highest degree that divides both f (X) and g(X), and two polynomials arerelatively prime if their gcd is 1.

In addition, since F[X ] is a Euclidean domain, the gcd d(X) of thepolynomials f (X) and g(X) may be expressed as a linear combination off (X) and g(X), i.e., there are polynomials a(X) and b(X) with d(X) =a(X) f (X) + b(X)g(X).

Furthermore, every Euclidean domain is a principal ideal domain (PID). Aprincipal ideal domain R is so named because it is an integral domain in which


every ideal I is principal, i.e., consists of the multiples of a single element i ,in which case we write I = 〈i〉. Applying that here, we see that every ideal ofF[X ] is of the form 〈 f (X)〉 for some polynomial f (X).

In an integral domain R, we have the notion of irreducible and prime ele-ments: A nonunit a ∈ R is irreducible if a = bc implies that b or c is a unit,and a nonunit a ∈ R is prime if a divides bc implies that a divides b or a di-vides c. In any integral domain R, every prime is irreducible, and in a PID R,every irreducible is prime, so in a PID R the notions of prime and irreducibleare the same.

Finally, in any PID unique factorization holds. In particular, it holds inF[X ]. Concretely, in F[X ] this means: Every nonconstant polynomial f (X) ∈F[X ] can be factored into irreducible polynomials in an essentially uniqueway. Furthermore, we may write f (X) uniquely (up to the order of the terms)as

f (X) = u f1(X) · · · fk(X) with u ∈ F, fi ∈ F[X ] monic irreducible

(and in particular, if f (X) is also monic then u = 1). We call this the primefactorization of f (X).

Given this background, we now get to work.

Lemma 2.2.1. Let f (X) ∈ F[X ] be a nonzero polynomial.(1) For a ∈ F, a is a root of f (X), i.e., f (a) = 0, if and only if X − a

divides f (X).(2) The number of roots of f (X) is less than or equal to the degree of

f (X).

Proof. (1) By the division algorithm, we may write f (X) = (X − a)q(X) +b, b ∈ F. Setting X = a, we see b = f (a).

(2) We prove this part of the lemma by induction on d = deg( f (X)),beginning with d = 0. If d = 0, then f (X) is a nonzero constant polynomial,so has 0 roots, and this part of the lemma holds. Now assume it holds for everypolynomial of degree d and let f (X) have degree d + 1. If f (X) has no roots,we are done. Otherwise, let f (α) = 0. Then f (X) = (X − α)g(X), so byinduction g(X) has e ≤ d roots. Then f (X) has e′ roots, with e′ = e or e + 1according as α is or is not a root of g(X), but in any case e′ ≤ d + 1 and weare done by induction. ��

Lemma 2.2.1 has the following important consequence.

Corollary 2.2.2. Let G be any finite subgroup of the multiplicative group of afield F. Then G is cyclic.

2.2 Polynomials 13

Proof. Let G have order n. Then any element g of G has order d for somed dividing n. Now an element g of G is a nonzero element of F, and so theequation gd = 1 in G is equivalent to the equation gd − 1 = 0 in F. Thuswe see that if g has order d, then g is a root of the polynomial Xd − 1. ByLemma 2.2.1, this polynomial has at most d roots. Hence G is an abeliangroup of order n with the property that G has at most d elements of order d,for any d dividing n. But it is easy to check that an abelian group with thisproperty must be cyclic (and furthermore that in this case G has exactly delements of order d , for any d dividing n). ��Lemma 2.2.3. Let F be a field and R an integral domain that is a finite-dimensional F-vector space. Then R is a field.

Proof. We need to show that any nonzero r ∈ R has an inverse. Consider{1, r, r2, . . . , }. This is an infinite set of elements of R, and by hypothesis Ris finite dimensional as an F-vector space, so this set is linearly dependent.Hence �n

i=0cir i = 0 for some n and some ci ∈ F not all zero. In other words,f (r) = 0 where f (X) = �n

i=0ci Xi ∈ F[X ].First, suppose c0 �= 0. Then

cnrn + · · · + c1r + c0 = 0,

c−10 (cnrn + · · · + c1r + c0) = 0,

c−10 cnrn + · · · + c−1

0 c1r + 1 = 0,

−c−10 cnrn − · · · − c−1

0 c1r = 1,

r(−c−10 cnrn−1 · · · − c−1

0 c1) = 1,

so r has inverse −c−10 cnrn−1 · · · − c−1

0 c1 ∈ R.If c0 = 0, write f (X) = Xkq(X) where the constant term of q(X) is

nonzero. Then 0 = f (r) = rkq(r) and R is an integral domain, so q(r) = 0and we may apply the above argument to the polynomial q(X). ��Proposition 2.2.4. Let f (X) ∈ F[X ] be an irreducible polynomial of degreed. Then F[X ]/〈 f (X)〉 is a field and is a d-dimensional F-vector space.

Proof. Let π : F[X ] → F[X ]/〈 f (X)〉 be the canonical projection and setπ( f (X)) = f (X).

By hypothesis, f (X) has degree d. We claim that S = {1, X , . . . , X d−1}is a basis for F[X ]/〈 f (X)〉.

S spans F[X ]/〈 f (X)〉: Let g(X) ∈ F[X ]/〈 f (X)〉. Then, by the divisionalgorithm, g(X) = f (X)q(X)+r(X) where either r(X) = 0 or deg r(X) < d.Then g(X) = r(X) = �d−1

i=0 ci X i is a linear combination of elements of S.


S is linearly independent: Suppose �d−1i=0 ci X i = g(X) = 0. Then g(X)

is divisible by f (X). But deg g(X) < deg f (X), so this is impossible unlessg(X) = 0, in which case c0 = c1 = . . . = cd−1 = 0.

Thus F[X ]/〈 f (X)〉 is a d-dimensional vector space over F.

Next observe that F[X ]/〈 f (X)〉 is an integral domain: Suppose g(X)h(X)

= 0. Then g(X)h(X) is divisible by f (X). Since f (X) ∈ F[X ] is irreducible,and hence prime, we have that f (X) divides g(X), in which case g(X) = 0,or f (X) divides h(X), in which case h(X) = 0.

Then, by Lemma 2.2.3, we conclude that F[X ]/〈 f (X)〉 is a field. ��Remark 2.2.5. It is worthwhile to give an explicit description of the field E =F[X ]/〈 f (X)〉. Let f (X) ∈ F[X ] be an irreducible polynomial of degree n,f (X) = an Xn +· · ·+a0. By taking representatives of equivalence classes, wemay regard

E = {g(X) ∈ F[X ] | deg g(X) < n}as a set. Addition in E is the usual addition of polynomials, while multipli-cation in E is defined as follows: Let g1(X), g2(X) ∈ F[X ] and let h(X) =g1(X)g2(X) ∈ F[X ]. We may write uniquely h(X) = f (X)q(X) + r(X)

where either r(X) = 0 or r(X) is a nonzero polynomial with deg r(X) <

deg f (X) = n. Then, for g1(X), g2(X) ∈ E,

g1(X)g2(X) = r(X) ∈ E,

where r(X) is defined as above.In particular, multiplication of polynomials g1(X), g2(X) ∈ E is the usual

multiplication of polynomials if deg g1(X) + deg g2(X) < n, while

Xn−1 · X = (−1/an)(an−1 Xn−1 + · · · + a0) ∈ E. �As we have remarked, we regard F ⊂ F[X ] by identifying the element a

of F with the constant polynomial f (X) = a. If π : F[X ] → F[X ]/〈 f (X)〉is the canonical projection, then π | F is an injection, and using this we alsoregard F ⊆ F[X ]/〈 f (X)〉.

In the proof of Proposition 2.2.4, we were careful to distinguish betweenf (X) ∈ F[X ] and f (X) = π( f (X)) ∈ F[X ]/〈 f (X)〉. But henceforth weshall follow the common practice of writing f (X) for π( f (X)) and relying onthe context to distinguish the two.

We now put these observations together to arrive at an important result.

Theorem 2.2.6 (Kronecker). Let f (X) ∈ F[X ] be any nonconstant polyno-mial. Then there is an extension field E ⊇ F in which f (X) has a root.

2.3 Extension Fields 15

Proof. Without loss of generality, we may assume that f (X) is irreducible.Set E = F[X ]/〈 f (X)〉. Then E is a field, E ⊇ F, and f (X) = 0 in E, so X isa root of f (X) in E. ��

Our last result in this section is a technical lemma that turns out to play animportant role in our development of Galois theory.

Lemma 2.2.7 (Invariance of GCD Under Field Extensions). Let E be anextension of F and let f (X) and g(X) be nonzero polynomials in F[X ]. Thenthe gcd of f (X) and g(X) as polynomials in E[X ] is equal to the gcd of f (X)

and g(X) as polynomials in F[X ]. In particular:(1) f (X) divides g(X) in F[X ] if and only if f (X) divides g(X) in E[X ].(2) f (X) and g(X) are relatively prime in F[X ] if and only if f (X) and

g(X) are relatively prime in E[X ].Proof. Let d(X) be the gcd of f (X) and g(X) in F[X ] and let d(X) be thegcd of F(X) and g(X) in E[X ]. Since d(X) is a polynomial in E[X ] dividingboth f (X) and g(X), d(X) divides d(X). On the other hand, since d(X) is thegcd of f (X) and g(X) in E[X ], there are polynomials a(X) and b(X) in E[X ]with

d(X) = f (X)a(X) + g(X)b(X).

Since d(X) divides both f (X) and g(X), d(X) divides d(X) as well.Hence d(X) and d(X) are two monic polynomials, each of which divides

the other, and so they are equal. ��(Actually, Lemma 2.2.7 (1) follows directly from our observation that the

quotient and remainder are independent of the field, and then Lemma 2.2.7 (1)can be used to prove the general case of Lemma 2.2.7 by considering Euclid’salgorithm.)

Remark 2.2.8. We have chosen to present a concrete proof of Proposition 2.2.4earlier in this section. But Proposition 2.2.4 is in fact a special case of a moregeneral result: If R is a PID and r ∈ R is an irreducible element, then R/〈r〉 isa field. This general result is proved in Appendix B. �

2.3 Extension Fields

In this section we study basic properties of field extensions.Let F, B, and E be fields with F ⊆ B ⊆ E. Then, as F-vector spaces, B is

a subspace of E, so (B/F) ≤ (E/F). But in this situation we can say more.


Lemma 2.3.1. Let F, B, and E be fields with F ⊆ B ⊆ E. Then

(E/F) = (E/B)(B/F).

Proof. Let {ε1, . . . , εm} be a basis for E as a B-vector space and {β1, . . . , βn}be a basis for B as an F-vector space. We claim that {εiβ j | i = 1, . . . , m, j =1, . . . , n} is a basis for E as an F-vector space, yielding the lemma.

First, we show that this set spans E. Let α ∈ E. Then α = �mi=1xiεi with

xi ∈ B. But for each i, xi = �nj=1 yi jβ j and so α = �m

i=1(�nj=1 yi jβ j )εi =

�yi j (εiβ j ).Next we show this set is linearly independent. Suppose �yi j (εiβ j ) = 0.

Then 0 = �yi j (εiβ j ) = �mi=1(�

nj=1 yi jβ j )εi . Since {εi } are linearly indepen-

dent, we have that, for each fixed i , �ni=1 yi jβ j = 0. Since {β j } are linearly

independent, we have that yi j = 0 for each j , completing the proof. ��Let E be an extension of F, and let B and D be subfields of E, both of which

are extensions of F. Then B ∩ D is a subfield of E which is also an extensionof F. Going in the opposite direction, we have the following definition.

Definition 2.3.2. Let E be an extension of F and let B and D be subfields of E,both of which are extensions of F. Then BD, the composite of B and D, is thesmallest subfield of E that contains both B and D, and is an extension of F. �Remark 2.3.3. (1) There is a smallest subfield as BD is the intersection of allsubfields of E that contain both B and D.(2) Clearly BD consists precisely of those elements of E of the form

(�mi=1bi di )(�

nj=1b′

j d′j )

−1

with bi , b′j ∈ B and di , d ′

j ∈ D. �There is an important case for which we do not need the denominators in

Remark 2.3.3 (2).

Lemma 2.3.4. Let E be an extension of F and let B and D be subfields of E,both of which are extensions of F. Suppose (D/F) is finite. Then (BD/B) ≤(D/F) and BD = {�m

i=1bi di | bi ∈ B, di ∈ D}. Furthermore, (BD/F) ≤(B/F)(D/F).

Proof. Let (BD)0 = {�mi=1bi di }. If {δi } is a basis for D as an F-vector space,

then {δi } spans (BD)0 as a B-vector space, so dimB(BD)0 ≤ dimF D. In par-ticular, if dimF D is finite, then dimB(BD)0 is finite, so (BD)0 is a field byLemma 2.2.3, and hence BD = (BD)0.

2.3 Extension Fields 17

Then, by Lemma 2.3.1,

(BD/F) = (BD/B)(B/F) ≤ (D/F)(B/F),

as claimed. ��One of the most common, and important, ways of obtaining field exten-

sions is by “adjoining” elements.

Definition 2.3.5. Let E be an extension of F, and let {αi } be a set of elementsof E. Then F({αi }) is the smallest subfield of E containing F and {αi }, and isthe field obtained by adjoining (or by the adjunction of) {αi }. �Remark 2.3.6. (1) Clearly F({αi }) is the set of elements of E that can be ex-pressed as rational functions of the elements of {αi } with coefficients in F.

(2) Suppose that {αi } = {α1, . . . , αn} is finite. Then

F(α1, . . . , αn) = F(α1)F(α2) · · · F(αn).

(3) Suppose that {αi } = {α1, . . . , αn} is finite. Then

F(α1, . . . , αn) = F(α1, . . . , αn−1)(αn). �Note that Remark 2.3.6 (3) says that in case {αi } is finite, we may obtain

F(α1, . . . , αn) by adjoining the elements of {αi } one at a time, and Remark2.3.6 (2) says that the order in which we adjoin them does not matter.

Note also that F(α) = F if (and only if) α ∈ F. Thus we think of obtainingF(α) by “adding in” α to F, and if α /∈ F, then this indeed gives us somethingnew, but if α ∈ F this does nothing.

Now that we know how to adjoin elements, let us give a concrete exampleof Lemma 2.3.4.

Example 2.3.7. (1) Let F = Q, B = Q(√

2), D = Q(√

3), and E = BD =Q(

√2,

√3). Then (BD/B) = 2 = (D/F) and (BD/F) = 4 = (B/F)(D/F).

(2) Let F = Q, B = Q(3√

2), D = Q(ω3√

2), and E = BD = Q(ω,3√

2).Here ω = (−1 + i

√3)/2 is a primitive cube root of 1. Note that ω is a root

of the quadratic polynomial X2 + X + 1. Observe that BD = B(ω). Then(BD/B) = 2 < (D/F) = 3 and (BD/F) = 6 < (B/F)(D/F) = 9.

Note that in both parts of this example B ∩ D = F. �Remark 2.3.8. From Lemma 2.3.4 we have that, since B and D are both exten-sions of B ∩ D, (BD/B ∩ D) ≤ (D/B ∩ D). Hence if B ∩ D ⊃ F, we certainlyhave that (BD/B ∩ D) < (D/F), and also that (BD/F) < (B/F)(D/F).

However, if B∩D = F, we may or may not have equality in Lemma 2.3.4,as Example 2.3.7 shows. There are many important cases in which we in-deed do have equality. We will investigate this question further in Section 3.4below. �


2.4 Algebraic Elements and Algebraic Extensions

Definition 2.4.1. (1) Let E be an extension of F. Then α ∈ E is algebraic overF if α is a root of some polynomial f (X) ∈ F[X ].

(2) E is an algebraic extension of F (or E/F is algebraic) if every α ∈ E isalgebraic over F. �

A complex number that is algebraic over Q is an algebraic number.

Lemma 2.4.2. The following are equivalent:(1) α is algebraic over F.(2) F(α)/F is finite.(3) F(α) = {polynomials in α with coefficients in F}.(4) α ∈ B for some finite extension B of F.

Proof. Note that in general F(α) = {rational functions in α with coefficientsin F}. If we let R = {polynomials in α with coefficients in F}, then R ⊆ F(α).

First, suppose (2) is true. Then R is a finite-dimensional vector space overF, so by Lemma 2.2.3, R is a field and R = F(α), proving (3). If (3) is true,then α−1 = f (α) is a polynomial in α, so α f (α) = 1. Then g(α) = 0 whereg(X) = X f (X)−1, and α is algebraic, proving (1). If (1) is true, then f (α) =0 for some polynomial f (X), so if f (X) has degree d, then {1, α, . . . , αd−1}spans R (as for any polynomial g(X), g(X) = f (X)q(X) + r(X) withdeg r(X) < d , and hence g(α) = f (α)q(α) + r(α) = r(α) = �d−1

i=0 ciαi );

so again R is a field, F(α) = R and (F(α)/F) ≤ d is finite, proving (2). Thuswe see that (1), (2), and (3) are equivalent. Also, (2) implies (4) since we maychoose B = F(α), and (4) implies (2) since if α ∈ B, then F(α) ⊆ B, so B/Ffinite implies F(α)/F finite. ��Lemma 2.4.3. Let α ∈ E be algebraic. Let

f1(X) be the monic generator of the ideal { f (X) ∈ F[X ] | f (α) = 0}f2(X) be the monic polynomial of lowest degree with f2(α) = 0.f3(X) be the monic irreducible polynomial with f3(α) = 0.

Then f1(X), f2(X), and f3(X) are well defined and f1(X) = f2(X) =f3(X). ��

The polynomial satisfying the equivalent conditions of this lemma plays avital role, and we give it a name.

Definition 2.4.4. Let α ∈ E be algebraic. The minimum polynomial mα(X) isthe polynomial satisfying the equivalent conditions of Lemma 2.4.3. �

2.4 Algebraic Elements and Algebraic Extensions 19

Remark 2.4.5. (1) Note that mα(X) depends on the field F, but in order to keepthe notation simple, we suppress that dependence.(2) Note that mα(X) is linear, in which case mα(X) = X − α, if and only ifα ∈ F. �

We have the following very important properties of mα(X):

Proposition 2.4.6. Let α be algebraic over F. Then(1) {1, α, α2, . . . , αd−1} is a basis for F(α) as a vector space over F, where

d = deg mα(X).(2) (F(α)/F) = deg mα(X).(3) Let T : F(α) → F(α) be multiplication by α, i.e., T (β) = αβ. Then

T is a linear transformation of F-vector spaces, and its minimum polynomialmT (X) = mα(X).

Proof. Let d = deg mα(X). Since mα(α) = 0, the set {1, α, . . . , αd−1} spansF(α), as in the proof of Lemma 2.4.2. We claim this set is linearly independent.Suppose not. Then �d−1

i=0 ciαi = 0 with not all ci = 0, i.e., f (α) = 0 where

0 �= f (X) = �d−1i=0 ci Xi ∈ F[X ]. But deg f (X) < deg mα(X), a contradiction

by Lemma 2.4.3. Hence {1, α, . . . , αd−1} is a basis for F(α), yielding (1), andthis basis has d elements, so (F(α)/F) = d, yielding (2).

(3) Let mα(X) = �di=0ci Xi . By definition, mα(X) is monic, so cd = 1.

Then 0 = �di=0ciα

i , so αd = �d−1i=0 (−ci )α

i .Now consider the matrix of T in the basis {1, α, . . . , αd−1} of F(α). Since

T (αi ) = αi+1, we see T has matrix⎡⎢⎢⎢⎢⎢⎣

0 0 ··· 0 −c0

1 0 ··· 0 −c1

0 1 ··· 0 −c2

......

......

0 0 ··· 1 −cd−1

⎤⎥⎥⎥⎥⎥⎦

which we recognize as the companion matrix C(mα(X)) of the polynomialmα(X). But for any polynomial f (X), C( f (X)) has minimum polynomial(and characteristic polynomial) f (X) (see [AW, Theorem 4.4.14]). ��Lemma 2.4.7. Let F ⊆ B ⊆ E and let α ∈ E. Then (B(α)/B) ≤ (F(α)/F).

Proof. If (F(α)/F) is infinite there is nothing to prove, so suppose (F(α)/F)

is finite. Then this follows immediately from Lemma 2.3.4. ��Corollary 2.4.8. Let α1, . . . , αn ∈ E with F(αi )/F finite for each i . ThenF(α1, . . . , αn)/F is finite.


Proof. Trivial for n = 1. For n = 2,

(F(α1, α2)/F(α1)) = ((F(α1)(α2)/F(α1)) ≤ (F(α2)/F)

by Lemma 2.3.4. Then proceed by induction. ��Corollary 2.4.9. (1) If E/F is finite, then E/F is algebraic.

(2) If E = F({αi }) with each αi algebraic over F, then E/F is algebraic.

Proof. (1) For any α ∈ E, F(α) ⊆ E, so (F(α)/F) ≤ (E/F) is finite, and so,by Lemma 2.4.2, E/F is algebraic.

(2) Let α ∈ E. Then α is a rational function of {αi }, so in particular α ∈F(α1, . . . , αn) for some n.

But, by Corollary 2.4.8, F(α1, . . . , αn)/F is finite, so, by part (1) andLemma 2.4.2, α is algebraic over F. Since α is arbitrary, E/F is algebraic. ��Remark 2.4.10. (1) It follows from Lemma 2.4.2 and Corollary 2.4.8 that ifE is an algebraic extension of F obtained by adjoining finitely many al-gebraic elements to F, then E is a finite extension of F. But not everyalgebraic extension E of F is a finite extension of F. For example, ifE = Q(

√2,

√3,

√5,

√7, . . . ), then E/Q is infinite. As a second example, if

E = Q(√

2,3√

2,4√

2,5√

2, . . . ), then E/Q is infinite. (While these two claimsare quite plausible, they of course require proof, but we shall not prove themnow.)

(2) We also remark that a complex number that is algebraic over Q isknown as an algebraic number, and that a subfield E of the field of com-plex numbers C that is obtained by adjoining finitely many algebraic num-bers to Q is known as an algebraic number field. We then immediately seethat E ⊆ C is an algebraic number field if and only if E is a finite extensionof Q. �

The next two results illustrate a theme which we will often see repeated.Let us consider fields F, B, and E with F ⊆ B ⊆ E. Thus B is an extensionof F and E is an extension of B. Also, E is an extension of F and B is asubextension. We have some “nice” property of extensions (in this case, beingalgebraic). We ask if a “nice” extension of a “nice” extension is necessarily a“nice” extension, and also if a subextension of a “nice” extension is necessarily“nice”. We will see that for some properties (e.g., being algebraic) the answersto both questions are yes, but for others (e.g., being Galois) they are not.

Proposition 2.4.11. Let E be an algebraic extension of F and let B be any fieldintermediate between F and E. Then B is an algebraic extension of F and E isan algebraic extension of B.

2.4 Algebraic Elements and Algebraic Extensions 21

Proof. Since E is an algebraic extension of F, every element of E is algebraicover F, and B ⊆ E, so certainly every element of B is algebraic over F, and soB is an algebraic extension of F.

Now let α ∈ E. Then α is a root of some polynomial f (X) ∈ F[X ], andF[X ] ⊆ B[X ], so α is certainly a root of some polynomial in B[X ], and so α

is algebraic over B. Since α is arbitrary, E is algebraic over B. ��Theorem 2.4.12. Let B be an algebraic extension of F and let E be an alge-braic extension of B. Then E is an algebraic extension of F.

Proof. Let α ∈ E. Then α is a root of some polynomial f (X) ∈ B[X ]. LetB0 ⊆ B be the field obtained from F by adjoining the finitely many coeffi-cients of f (X), each of which is algebraic over F. Then (B0/F) is finite and(B0(α)/B0) is finite, so (B0(α)/F) = (B0(α)/B0)(B0/F) is finite as well.But α ∈ B0(α), so α is an element of a finite extension of F and hence,by Lemma 2.4.2, α is algebraic over F. Since α is arbitrary, E is algebraicover F. ��

For any fixed element α ∈ E, we may define a ring homomorphismϕ : F[X ] → E by ϕ | F = id and ϕ(X) = α (so that ϕ( f (X)) = f (α)

for every f (X) ∈ F[X ]).Lemma 2.4.13. Let α ∈ E be algebraic over F. Then ϕ induces

ϕ : F[X ]/〈mα(X)〉 → F(α)

an isomorphism of fields and of F-vector spaces.

Proof. Let π : F[X ] → F[X ]/〈mα(X)〉 be the canonical projection and setπ( f (X)) = f (X).

First, we show that ϕ is well defined. Let f (X) ∈ F[X ]/〈mα(X)〉 andsuppose that π( f (X)) = f (X) = π(g(X)). Then g(X)− f (X) is divisible bymα(X), so g(X) = f (X) + mα(X)q(X) for some polynomial q(X) ∈ F[X ],and then ϕ(g(X)) = g(α) = f (α) + mα(α)q(α) = f (α) = ϕ( f (X)), asrequired.

Now it is easy to check that ϕ is a homomorphism of rings and a linearmap of F-vector spaces, so to complete the proof of the lemma we need onlyshow that it is both one-to-one and onto.

Since ϕ(1) = 1, we see, by Lemma 2.1.3, that ϕ is injective.Now, by Proposition 2.2.4 and Proposition 2.4.6 (2), F[X ]/〈mα(X)〉 and

F(α) are both F-vector spaces of the same finite dimension, and hence, sinceϕ is injective, it is surjective as well. ��


Remark 2.4.14. We have chosen to present a direct proof of Lemma 2.4.13, butit is a special case of a more general algebraic result: Let ϕ : R → S be a mapof rings. Then ϕ induces an isomorphism ϕ : R/ Ker(ϕ) → Im(ϕ). �Remark 2.4.15. Let f (X) ∈ F[X ] be a monic irreducible polynomial, and letE be an extension field in which f (X) has a root α. Then f (X) = mα(X) isthe minimum polynomial of α. Of course, F(α) ⊆ E, and F(α) = E if andonly if (E/F) = deg mα(X), by Proposition 2.4.6 (2). �

2.5 Splitting Fields

Definition 2.5.1. (1) Let f (X) ∈ F[X ] be a nonconstant polynomial. If E ⊇ Fis such that f (X) factors into (possibly repeated) linear factors in E[X ], thenf (X) splits in E[X ].

(2) If f (X) splits in E[X ] but not in B[X ] for any proper subfield B of E,then E is a splitting field of f (X). �

Note that X − α is a factor of f (X) if and only if f (α) = 0, i.e., if andonly if α is a root of f (X). Thus, we regard f (X) as splitting in E if “all” theroots of f (X) are in E, and we regard E as a splitting field if “all” the roots off (X) lie in E but not in any proper subfield of E.

Lemma 2.5.2. Let f (X) ∈ F[X ] be a nonconstant polynomial. Then f (X)

has a splitting field.

Proof. By induction on deg f (X), and all fields simultaneously. Without lossof generality we may assume that f (X) is monic. If deg f (X) = 1, thenf (X) = X − a ∈ F[X ] and F is a splitting field for f (X).

Assume the lemma holds for all polynomials of degree less than d, and allfields, and let f (X) ∈ F[X ] have degree d.

By Theorem 2.2.6, there is an extension field B of F in which f (X) has aroot α1, and moreover B = F(α1). Then f (X) = (X − α1)g(X) ∈ B[X ]. Byinduction, g(X) has a splitting field E over B. Hence g(X) = (X−α2) · · · (X−αd) ∈ E[X ] and f (X) = (X − α1) · · · (X − αd) ∈ E[X ] splits.

Clearly E ⊇ B(α2, . . . , αn), and then E = B(α2, . . . , αn), since certainlyg(X) does not split over any field not containing each of α2, . . . , αn . ThenE = F(α1)(α2, . . . , αn) = F(α1, . . . , αn), and f (X) does not split in anyproper subfield of E, by the same logic, so E is a splitting field for f (X). ��

Note that in general it may well be the case that an extension B of F con-tains a root of an irreducible polynomial f (X) ∈ F[X ] but that polynomial

2.5 Splitting Fields 23

does not split in B[X ], i.e., that “some” but not “all” of the roots of f (X) arein B. As an example of this, we may take B = Q(

4√

2). Then the polynomialX4 − 2 is irreducible in Q[X ] with roots 4

√2, i 4

√2,− 4

√2, and −i 4

√2, and B

contains just two of these four roots ( 4√

2 and − 4√

2).We shall soon see that any two splitting fields of f (X) ∈ F[X ] are isomor-

phic. Granting that, for the moment, we have the following result.

Lemma 2.5.3. Let f (X) ∈ F[X ] be a nonconstant polynomial, with deg f (X)

= d. Let E be a splitting field of f (X). If f (X) is irreducible, (E/F) is divisi-ble by d. In any case, (E/F) ≤ d!.Proof. Without loss of generality we may assume that f (X) is monic.

Suppose that f (X) is irreducible, and let α ∈ E be a root of f (X). Thenf (X) = mα(X), by Lemma 2.4.3, and d = deg mα(X) = (F(α)/F), byProposition 2.4.6 (2). But F(α) ⊆ E, and so (F(α)/F) divides (E/F).

For the upper bound, we proceed by induction on d, and all fields simul-taneously. If d = 1, E = F and (E/F) = 1. Suppose the lemma holds forall polynomials of degree less than d, and all fields, and let f (X) ∈ F[X ]have degree d. Let α be a root of f (X), set B = F(α), and write f (X) =(X − α)g(X) with g(X) ∈ B[X ]. Since f (α) = 0, mα(X) divides f (X), so(B/F) = deg mα(X) ≤ deg f (X) = d .

Let E be a splitting field of g(X), a polynomial of degree d − 1, over B.Then, by induction, (E/B) ≤ (d − 1)!, and then (E/F) = (E/B)(B/F) ≤ d!.

��Example 2.5.4. (1) Let F = Q and let E be the splitting field of the irreduciblepolynomial X3 − 2 ∈ F[X ]. Then E = F(ω,

3√

2) where ω2 + ω + 1 = 0, and(E/F) = 6 = 3!.

(2) Let F = Q(ω) and let E be the splitting field of the irreducible polyno-mial X3 − 2 ∈ F[X ]. Then E = F(

3√

2), and (E/F) = 3 < 3!. �Corollary 2.5.5. Let E be a splitting field of a nonconstant polynomial f (X) ∈F[X ]. Then E/F is algebraic.

Proof. E/F is finite. ��Remark 2.5.6. Suppose that f (X) ∈ F[X ] splits in some extension A of F, andlet α1, . . . , αk be the roots of f (X) in A (so that f (X) = (X − α1)

d1 · · · (X −αk)

dk ∈ A[X ]). Then E = F(α1, . . . , αk) is a splitting field for f (X), asf (X) visibly splits over E, and any field over which f (X) splits must containα1, . . . , αk , and hence must contain E. �


2.6 Extending Isomorphisms

We now prove two results about extensions of field isomorphisms that will bekey technical tools.

Lemma 2.6.1. Let σ0 : F1 → F2 be an isomorphism of fields. Let f1(X) ∈F1[X ] be irreducible and let E1 = F1(β1) where f1(β1) = 0. Let f2(X) =σ0( f1(X)) and let E2 = F2(β2) where f2(β2) = 0. Then σ0 extends to aunique isomorphism σ : E1 → E2 with σ(β1) = β2.

Proof. Clearly σ0 gives an isomorphism σ0 : F1[X ] → F2[X ] and f2(X) isirreducible if and only if f1(X) is. Now, by Lemma 2.4.13, we have isomor-phisms

ϕi : Fi [X ]/〈 fi (X)〉 → Fi (βi ) = Ei , i = 1, 2,

and we let

σ = ϕ2σ0ϕ−11 . ��

Corollary 2.6.2. Let f (X) ∈ F[X ] be irreducible. Suppose β1 and β2 areboth roots of f (X) and let Ei = F(β), i = 1, 2. Then there is a uniqueisomorphism σ : E1 → E2 with σ | F = id and σ(β1) = β2. In particular, ifE = F(β1) = F(β2), there is a unique automorphism σ of E with σ | F = idand σ (β1) = β2. ��

This corollary says that, algebraically, there is no distinction between rootsof the same irreducible polynomial — choosing any two roots gives us isomor-phic extensions.

Lemma 2.6.3. Let σ0 : F1 → F2 be an isomorphism of fields. Let f1(X) ∈F1[X ] and let f2(X) = σ0( f1(X)) ∈ F2[X ]. Let E1 be a splitting field of f1(X)

and let E2 be a splitting field of f2(X). Then σ0 extends to an isomorphismσ : E1 → E2.

Proof. Factor f1(X) into irreducibles in F[X ], and let there be k factors. SetdF( f1) = deg( f1) − k. We prove the theorem by induction on dF( f1), and allfields.

If dF( f1) = 0, then f1(X) is a product of linear factors, and E1 = F1,E2 = F2, so σ = σ0.

Suppose dF( f1) > 0. Then f1(X) has an irreducible factor g1(X) of de-gree greater than 1. Let α1 be a root of g1(X) in E1 and let α2 be a root ofσ0(g1(X)) in E2. Then E1 ⊇ F(α1) and E2 ⊇ F(α2). By Lemma 2.6.1, thereis an isomorphism σ1 : F(σ1) → F(σ2) with σ1 | F = σ0 and σ1(α1) = α2.

2.7 Normal, Separable, and Galois Extensions 25

Set B = F(α1), and consider f1(X) ∈ B[X ]. Now g1(X) ∈ B[X ] has thefactor X − α1, so f1(X) has at least k + 1 irreducible factors in B[X ] andhence dB( f1) < dF( f1). Since E1 was a splitting field of f1(X), regarded aspolynomial in F1[X ], and B ⊆ E1, then, by Remark 2.5.6, E1 is still a splittingfield of f1(X) regarded as a polynomial in B[X ], and similarly for E2, so byinduction σ1 extends to σ , as claimed. ��Corollary 2.6.4. Let f (X) ∈ F[X ]. Then any two splitting fields of f (X) areisomorphic. ��

Note that we are certainly not claiming that the isomorphism σ of Lemma2.6.3 is unique! In fact, studying automorphisms E that fix F is what Galoistheory is all about. To give a simple example of this, let F = Q and E =Q(

√D). Then E has an automorphism σ given by σ(a + b

√D) = a − b

√D.

2.7 Normal, Separable, and Galois Extensions

We begin by defining two (independent) properties of field extensions.

Definition 2.7.1. An algebraic extension E of F is a normal extension if everyirreducible polynomial f (X) ∈ F[X ] with f (α) = 0 for some α ∈ E splits inE[X ]. �Remark 2.7.2. Observe that Definition 2.7.1 is equivalent to: An algebraic ex-tension E of F is normal if mα(X) splits in E[X ] for every α ∈ E. �Definition 2.7.3. (1) If the irreducible factors of f (X) ∈ F[X ] split into aproduct of distinct linear factors in some (hence in any) splitting field forf (X), then f (X) is a separable polynomial. Otherwise, it is an inseparablepolynomial.

(2) Let E be an algebraic extension of F. Then α ∈ E is a separable el-ement if mα(X) is a separable polynomial. Otherwise, α is an inseparableelement.

(3) Let E be an algebraic extension of F. Then E is a separable extensionof F if every α ∈ E is a separable element. Otherwise, it is an inseparableextension. �Remark 2.7.4. We are most interested in extensions that are both normal andseparable, as these have the closest connection with Galois theory, so we willdefer our consideration of extensions that satisfy only one of these two prop-erties until later. However, some discussion of both of these properties is inorder now.


First, we might observe that the definitions of normal and separable arerather discouraging on the face of it, as they each require us to verify a condi-tion on mα(X) for each α ∈ E. But, as we will soon see (in Theorem 2.7.14),we have an easy criterion for extensions to be both normal and separable.

Even if we are interested in extensions that are normal or separable, thesituation is not so bad. We note some results that we shall prove later on.

Theorem 5.2.1. E is a finite normal extension of F if and only if E isthe splitting field of a polynomial f (X) ∈ F[X ].Theorem 5.1.9. E is a finite separable extension of F if and only ifE is obtained from F by adjoining root(s) of a separable polynomialf (X) ∈ F[X ].

Actually, sometimes separability is automatic. There are some fields F allof whose algebraic extensions are separable. Such a field F is called perfect,and we will show below that the following is true.

Theorem 3.2.6. F is a perfect field if and only if(1) char(F) = 0, or(2) char(F) = p and F = Fp.

Corollary 3.2.7. Every finite field is perfect. �We now define one of the most important objects in mathematics.

Definition 2.7.5. Let E be an algebraic extension of F. The Galois groupGal(E/F) is the group of all automorphisms of E that restrict to the identityon F,

Gal(E/F) = {σ : E → E automorphism | σ | F = id}. �(Strictly speaking, we should have defined Gal(E/F) to be the set of these

automorphisms, but it is easy to verify that this set is indeed a group undercomposition.)

Lemma 2.7.6. Let E be an algebraic extension of F and let B be a field withF ⊆ B ⊆ E. Then Gal(E/B) is a subgroup of Gal(E/F).

Proof. Any automorphism σ : E → E with σ | B = id certainly satisfiesσ | F = id. ��Definition 2.7.7. Let G be a group of automorphisms of a field E. Then thefixed field Fix(G) = {α ∈ E | σ(α) = α for all σ ∈ G}. �

2.7 Normal, Separable, and Galois Extensions 27

Lemma 2.7.8. Let G be a group of automorphisms of a field E. Then Fix(G)

is a subfield of E. ��Note that, by definition, F ⊆ Fix(Gal(E/F)).

Definition 2.7.9. Let E be an algebraic extension of F. Then E is a Galoisextension of F if Fix(Gal(E/F)) = F. �

Our first main goal is the Fundamental Theorem of Galois Theory (FTGT),which, for a finite Galois extension E of F, establishes a one-to-one correspon-dence between intermediate fields B (i.e., fields B with F ⊆ B ⊆ E) and sub-groups H of G = Gal(E/F). On the way there, we need to obtain a criterionfor E to be a Galois extension of F, an important result in its own right.

Definition 2.7.10. Let E be an algebraic extension of F and let α ∈ E. Then{σ(α) | σ ∈ Gal(E/F)} is the set of (Galois) conjugates of α in E. �Lemma 2.7.11. Let H = {σ ∈ Gal(E/F) | σ(α) = α}. Then H is a subgroupof G = Gal(E/F) and the number of conjugates of α in E is equal to the index[G : H ]. ��

The following lemma is particularly useful.

Lemma 2.7.12. Let E be a Galois extension of F and let α ∈ E. Then α hasfinitely many conjugates in E. If {αi }i=1,...,r is the set of conjugates of α = α1,then

mα(X) =r∏

i=1

(X − αi ).

Proof. Since mα(X) ∈ F[X ], for each σ ∈ G = Gal(E/F), σ(α) is a rootof σ(mα(X)) = mα(X). Since mα(X) has only finitely many roots, {σ(α)}must be a finite set. Now let nα(X) = ∏r

i=1(X − αi ). Since σ ∈ G permutes{αi }, σ(nα(X)) = nα(X). In other words, every σ ∈ G fixes each coefficientof nα(X). Since E is a Galois extension of F, this implies that each of thesecoefficients is in F, i.e., that nα(X) ∈ F[X ]. As we have observed, mα(αi ) = 0for each i , so X−αi divides mα(X) for each i , and hence nα(X) divides mα(X)

in E[X ], and hence in F[X ], by Lemma 2.2.7. But mα(X) is irreducible inF[X ], so mα(X) = nα(X). ��Corollary 2.7.13. Let E be a Galois extension of F and let α ∈ E. Then(F(α)/F) is equal to the number of conjugates of α in E.

Proof. (F(α)/F) = deg mα(X) by Proposition 2.4.6 (2) and deg mα(X) isequal to the number of conjugates of α in E by Lemma 2.7.12. ��


The following theorem is a key result.

Theorem 2.7.14. Let E be a finite extension of F. The following are equiva-lent:

(1) E is a Galois extension of F.(2) E is a normal and separable extension of F.(3) E is the splitting field of a separable polynomial f (X) ∈ F[X ].

Proof. (1) ⇒ (2): Let E be a Galois extension of F and let α ∈ E. Let {αi } bethe set of conjugates of α in E. By Lemma 2.7.12, mα(X) = ∏r

i=1(X − αi ),and this is a separable polynomial (as its roots are distinct) that splits in E.

(2) ⇒ (3): Let {εi } be a basis for E over F, and let f (X) be the productof the distinct elements of {mεi (X)}. Then f (X) is separable, as each mεi (X)

has distinct roots, and f (X) certainly splits in E. Moreover, any field in whichf (X) splits must contain each εi , so E is the splitting field of f (X).

(3) ⇒ (1): We prove this by induction on (E/F) and all fields simultane-ously. Let E be the splitting field of the separable polynomial f (X).

If (E/F) = 1, then E = F, Gal(E/F) = {id}, and we are done.Suppose now that the theorem is true for all extensions of degree less than

n and let (E/F) = n > 1.Factor f (X) into irreducibles f (X) = f1(X) · · · fk(X) in F[X ]. Some

fi (X) has degree greater than one, as otherwise f (X) would split in F[X ]and so E = F. Renumbering if necessary, we may suppose that this factor isf1(X). Let it have degree r . Let α = α1 be a root of f1(X), α ∈ E. Thenf1(X) = mα(X), since f1(X) is irreducible. Of course, α /∈ F.

By assumption, mα(X) is separable, so has distinct roots α1, . . . , αr . Nowαi is a root of the irreducible polynomial mα(X) by assumption, and αi is aroot of the irreducible polynomial mαi (X) by definition, so, by Lemma 2.4.3,mαi (X) = mα(X), for each i . Furthermore, by Corollary 2.6.2, there are iso-morphisms σi : F(α1) → F(αi ) with σi | F = id and σi (α1) = αi , i =1, . . . , r .

Consider E ⊇ F(αi ) ⊃ F. Then E is the splitting field of mα(X) ∈F(αi )[X ], so, by Lemma 2.6.3, there is an automorphism of E extending theisomorphism σi . Denote this extension by σi as well. Then σi | F = id, soσi ∈ Gal(E/F). Note that σi (F(α j )) = F(αk) for some k, as the automor-phism σi of E must permute the roots of the irreducible polynomial mα(X).

Now consider E as an extension of F(α). Then (E/F(α)) < (E/F), andE is the splitting field of the separable polynomial mα(X) ∈ F(α)[X ], soby induction E is a Galois extension of F(α), i.e., F(α) is the fixed field ofGal(E/F(α)). By Lemma 2.7.6, Gal(E/F(α)) is a subgroup of Gal(E/F), sothis implies that B = Fix(Gal(E/F)) ⊆ F(α). We wish to show that B = F.

2.8 The Fundamental Theorem of Galois Theory 29

Let mα(X) ∈ B[X ] be the minimum polynomial of α regarded as an ele-ment of B. Since B ⊆ F(α), we see that B(α) = F(α), and then

deg mα(X) = (B(α)/B) = (F(α)/B) ≤ (F(α)/F) = deg mα(X)

with equality if and only if B = F. Thus it remains to show that deg mα(X) =deg mα(X). In fact we show that mα(X) = mα(X).

Now mα(X) divides mα(X) and mα(X) = ∏ri=1(X − αi ) by Lemma

2.7.12.Consider αi . We have constructed an element σi ∈ Gal(E/F) with σi (α) =

αi , i.e., there is an automorphism σi of E with σi (α) = αi and σi | F = id.But B = Fix(Gal(E/F)) so σi | B = id, and hence σi ∈ Gal(E/B). Thenσi (mα(X)) = mα(X), as σi (mα(X)) ∈ B[X ]; thus

mα(αi ) = mα(σi (α)) = σi (mα(α)) = σi (0) = 0,

so αi is a root of mα(X) for each i , i.e., X − αi is a factor of mα(X) for eachi , and so mα(X) = mα(X), as required. ��

2.8 The Fundamental Theorem of Galois Theory

In this section we reach our first main goal. Before we get there, we have somework to do.

We let F∗ denote the set of nonzero elements of F, and recall that F∗ is agroup under multiplication.

Definition 2.8.1. A character of a group G in a field F is a homomorphismσ : G → F∗. �Example 2.8.2. (1) Let G = F∗. Then id : G → F∗ is a character.

(2) Let G = F∗. For any automorphism σ of F, σ : G → F∗ is a character.(3) Let G = F∗. For any map of fields σ : F → E, σ : G → E∗ is a

character. �Definition 2.8.3. A set of characters {σi }i=1,...,n is dependent if there is a set ofelements {ai }i=1,...,n of F, not all zero, with a1σ1 + · · · + anσn = 0 (i.e., with(a1σ1 +· · ·+ anσn)(x) = 0 for every x ∈ G). Otherwise it is independent. �Theorem 2.8.4 (Dirichlet). Let {σi }i=1,...,n be a set of mutually distinct char-acters of G in F. Then {σi } is independent.


Proof (Artin). By induction on n.Suppose n = 1. Then a1σ1 = 0 implies a1 = 0 (as σ(1) = 1).Suppose now that the theorem is true for any set of fewer than n characters

and consider {σi }i=1,...,n . Suppose a1σ1 + · · · + anσn = 0. Some ai is nonzero.Let that be a1. Since σ1 and σn are distinct, there is an x ∈ G with σ1(x) �=σn(x). Now by hypothesis, for every y ∈ G,

a1σ1(y) + · · · + anσn(y) = 0,

so, multiplying by σn(x),

(∗) a1σn(x)σ1(y) + · · · + anσn(x)σn(y) = 0.

On the other hand, for every y ∈ G,

a1σ1(xy) + · · · + anσn(xy) = 0;using the fact that σi (xy) = σi (x)σi (y),

(∗∗) a1σ1(x)σ1(y) + · · · + anσn(x)σn(y) = 0,

and then, subtracting (∗) from (∗∗),

a1(σ1(x) − σn(x))σ1(y) + · · · + an−1(σn−1(x) − σn(x))σn−1(y) = 0

for every y ∈ G. By induction, each coefficient is zero, so in particular σ1(x) =σn(x), a contradiction. ��Theorem 2.8.5. Let � = {σi , i = 1, . . . , n} be a group of mutually distinctautomorphisms of a field E, and let F be the fixed field of �. Then (E/F) = n.

Proof. (Artin) First we show n ≤ (E/F) (which only uses the fact that � is aset) and then we show n ≥ (E/F) (which uses the fact that � is a group). Weprove both inequalities by contradiction. Let r = (E/F).

For the first step, suppose r < n. Let {ε1, . . . , εr } be a basis for E over F.Consider the system of equations in E:

⎡⎢⎣

σ1(ε1) · · · σn(ε1)...

σ1(εr ) · · · σn(εr )

⎤⎥⎦

⎡⎢⎣

α1...

αn

⎤⎥⎦ = 0.

This system has more unknowns than equations, so it has a nontrivial solutionwhich we still denote by α1, . . . , αn . Then


n∑i=1

αiσi (εk) = 0 for each k = 1, . . . , r.

Let ε ∈ E be arbitrary. Since {εk} is a basis, we have that ε = ∑rk=1 fkεk with

fk ∈ F, and since F is the fixed field of �, σi ( fk) = fk for each i, k. Thenn∑

i=1

αiσi (ε) =n∑

i=1

αiσi (

r∑k=1

fkεk) =∑i,k

αiσi ( fkεk) =∑i,k

αi fkσi (εk)

=r∑

k=1

fk(

n∑i=1

αiσi (εk)) =r∑

k=1

fk(0) = 0,

so∑n

i=1 αiσi = 0, contradicting Theorem 2.8.4.

For the second step, suppose r > n. Since � is a group, some σi is theidentity. Let it be σ1. Let {ε1, . . . , εn+1} be a set of F-linearly independentelements of E. Consider the system of equations in E:⎡

⎢⎣σ1(ε1) · · · σ1(εn+1)

...

σn(ε1) · · · σn(εn+1)

⎤⎥⎦

⎡⎢⎣

β1...

βn+1

⎤⎥⎦ = 0.

This system has more unknowns than equations, so it has a nontrivial solu-tion which we still denote by β1, . . . , βn+1. Among all such solutions chooseone with the smallest number s of nonzero values. By permuting the subscripts,we may assume this solution is β1, . . . , βs, 0, . . . , 0. Note that s > 1, since ifs = 1, 0 = β1σ1(ε1) = β1ε1 implies β1 = 0 as ε1 �= 0. Then we may fur-ther assume (multiplying by β−1

s if necessary) that βs = 1. Finally, not all βi

are in F, since if they were, we would have 0 = β1σ1(ε1) + · · · + βsσ1(εs) =β1ε1 +· · ·+βsεs , an F-linear dependence of {ε1, . . . , εn}. Thus we may finallyassume that β1 /∈ F. Then we have

(∗i ) β1σi (ε1) + · · · + βs−1σi (εs−1) + σi (εs) = 0, i = 1, . . . , n.

Since β1 /∈ F = Fix(�), there is a σk ∈ � with σk(β1) �= β1. Since � is agroup, for any σi ∈ � there is a σ j ∈ � with σi = σkσ j . Then applying σk to(∗ j ), we have

(∗∗i ) σk(β1)σi (ε1) + · · · + σk(βs−1)σi (εs−1) + σi (εs) = 0,

for i = 1, . . . , n. Subtracting (∗∗i ) from (∗i ) gives

(∗∗∗i ) (β1 − σk(β1))σi (ε1) + · · · + (βs−1 − σk(εs−1))σi (εs−1) = 0,

for i = 1, . . . , n. Now β1 − σk(β1) �= 0 and this equation has fewer than snonzero coefficients, a contradiction. ��


Corollary 2.8.6. (1) Let G be a finite group of automorphisms of E and letF = Fix(G). Then any automorphism of E fixing F must belong to G.

(2) Let G1 and G2 be two distinct finite groups of automorphisms of E, andlet F1 = Fix(G1), F2 = Fix(G2). Then F1 and F2 are distinct.

Proof. (1) (E/F) = |G| = n, say, by Theorem 2.8.5, so if there were someautomorphism σ of E fixing F that was not in G, then F would be fixed by atleast n + 1 > (E/F) automorphisms of E, a contradiction.

(2) We prove the contrapositive: If F1 = F2, then F1 is fixed by G2, soG2 ⊆ G1 by part (1), and vice versa, so G1 = G2. ��Lemma 2.8.7. Let G be a group of automorphisms of E fixing F and let H bea subgroup of G. Let B = Fix(H). Then for any σ ∈ G, σ(B) = Fix(σ Hσ−1).

Proof. Let β ∈ B and τ ∈ H . Then στσ−1(σ (β)) = στ(β) = σ(β), soσ(B) ⊆ Fix(σ Hσ−1). On the other hand, if στσ−1(β ′) = β ′ for some β ′ ∈ Eand all τ ∈ H , then τσ−1(β ′) = σ−1(β ′); so σ−1(β ′) = β ∈ Fix(H) = B andβ ′ = σ(β), so Fix(σ Hσ−1) ⊆ σ(B). ��

Here is the most important theorem in this book, and one of the fundamen-tal theorems in all of mathematics.

Theorem 2.8.8 (Fundamental Theorem of Galois Theory = FTGT). Let Ebe a finite Galois extension of F and let G = Gal(E/F).

(1) There is a one-to-one correspondence between intermediate fieldsE ⊇ B ⊇ F and subgroups {1} ⊆ GB ⊆ G given by

B = Fix(GB).

(2) B is a normal extension of F if and only if GB is a normal subgroup ofG. This is the case if and only if B is a Galois extension of F. In this case

Gal(B/F) ∼= G/GB.

(3) For each E ⊇ B ⊇ F, (B/F) = [G : GB] and (E/B) = |GB|.We say that the intermediate field B and the subgroup GB “belong” to each

other.Before proving this theorem, we make some observations.

Remark 2.8.9. (1) If B1 ⊆ B2, then GB1 ⊇ GB2 , so the correspondence is“inclusion-reversing”.

(2) E is always a Galois extension of B and GB = Gal(E/B), both claimsbeing true by the definition of a Galois extension (Definition 2.7.9).


(3) Since E/F is separable, B/F is separable for any B intermediate be-tween F and E (recall Definition 2.8.3 (3)), so, by Theorem 2.7.14, B/F isnormal if and only if B/F is Galois. This justifies the second sentence ofTheorem 2.8.8 (2).

(4) As we shall see from the proof, in case B is a Galois extension ofF, the quotient map Gal(E/F) → Gal(B/F) is simply given by restriction,σ �→ σ | B. �Proof. (1) For each subgroup H and G, let

BH = Fix(H).

This gives a map

� : {subgroups of G} → {fields intermediate between F and E}.We show � is a one-to-one correspondence.� is one-to-one: If H1 �= H2, then Fix(H1) �= Fix(H2) by Corollary

2.8.6 (2).� is onto: Let F ⊆ B ⊆ E and let

H = {σ ∈ G | σ | B = id} = Gal(E/B) ⊆ Gal(E/F).

Since E/F is Galois, it is the splitting field of a separable polynomialf (X) ∈ F[X ] by Theorem 2.7.14. But F ⊆ B so f (X) ∈ B[X ]. Then Eis the splitting field of the separable polynomial f (X) ∈ B[X ], so E/B isGalois, again by Theorem 2.7.14, and hence B = Fix(Gal(E/B)) = Fix(H).

Thus � is one-to-one and onto, and if �−1(B) = GB, then B = Fix(GB).

(3) Since E/B is Galois, we know that (E/B) = | Gal(E/B)| = |GB|, byTheorem 2.8.5, and (E/F) = (E/B)(B/F), while |G| = |GB|[G : GB], so(B/F) = [G : GB].

(2) Suppose that GB is a normal subgroup of G. For any σ ∈ Gal(E/F),σ(B) = Fix(σ GBσ−1) = Fix(GB) = B, by Lemma 2.8.7. Hence we havethe restriction map R : Gal(E/F) → Gal(B/F) given by σ �→ σ | B,and Ker(R) = {σ ∈ G | σ | B = id} = GB by definition. ThusIm(R) ∼= Gal(E/F)/ Ker(R) = G/GB. But let σ0 ∈ Gal(B/F). Then E/Bis Galois, so E is the splitting field of a (separable) polynomial f (X). Thenσ0 : B → B extends to σ : E → E by Lemma 2.6.3, and σ | F = id, soσ ∈ Gal(E/F). Then R(σ ) = σ0, so R : Gal(E/F) → Gal(B/F) is onto andhence Gal(B/F) ∼= G/GB.


Conversely, suppose that B/F is Galois. Then B is the splitting field of aseparable polynomial f (X). Let f (X) have roots β1, . . . , βr in B, and note,by Remark 2.5.6, that B = F(β1, . . . , βr ). Let σ ∈ G = Gal(E/F).

Now f (X) ∈ F[X ], so σ( f (X)) = f (X) and hence σ permutes theroots of f (X), i.e., for every i , σ(βi ) = β j for some j . In particular thisimplies that σ(B) = B. But now, by Lemma 2.8.7, Fix(GB) = B = σ(B) =Fix(σ GBσ−1), so, by part (1), GB = σ GBσ−1. Since σ is arbitrary, GB is anormal subgroup of G. ��

In the next section, we shall give a number of examples of the FTGT, butfirst we continue here with our theoretical development.

Definition 2.8.10. Two fields B1 and B2 intermediate between E and B areconjugate if there is a σ ∈ Gal(E/F) with σ(B1) = B2. �Corollary 2.8.11. (1) Two intermediate fields B1 and B2 are conjugate if andonly if GB1 and GB2 are conjugate subgroups of Gal(E/F).

(2) B is its only conjugate if and only if GB is a normal subgroup ofGal(E/F).

Proof. Immediate from Lemma 2.8.7 and Theorem 2.8.8. ��Returning to the situation of the Fundamental Theorem of Galois Theory,

let E be a finite Galois extension of F, and let B be an intermediate field be-tween E and F.

Definition 2.8.12. Let

QB = {σ0 : B → B′ an isomorphism | F ⊆ B′ ⊆ E, σ0 | F = id}.Let � : QB → {left cosets of GB in G} be defined as follows:Let σ0 ∈ QB. Since E is a splitting field, σ0 extends to σ ∈ Gal(E/F) by

Lemma 2.6.3. Then set

�(σ0) = [σ ] ∈ G/GB. �Proposition 2.8.13. (1) � is well defined.

(2) � is one-to-one and onto.(3) Gal(B/F) ⊆ QB with equality if and only if GB is normal, or equivalent

if and only if B/F is Galois. In this case, � is an isomorphism of groups,� : QB → G/GB.

Proof. (1) We need to check that �(σ0) is independent of the choice of exten-sion σ . Let σ ′ be another extension of σ0. Then σ ′ | B = σ0 = σ | B, i.e.,σ ′(β) = σ(β) for all β ∈ B, so (σ−1σ ′)(β) = β for all β ∈ B, and henceσ−1σ ′ ∈ GB, so σ ′ ∈ σ GB and [σ ′] = [σ ].


(2) Suppose �(σ0) = [σ ] and �(τ0) = [τ ] with [σ ] = [τ ]. Then τ = σρ

for some ρ ∈ GB, so τ | B = σρ | B. But ρ ∈ GB so ρ | B = id; henceτ | B = σ | B, i.e., τ0 = σ0, and � is one-to-one. Also, let σ GB be a leftcoset of GB in G, and let σ0 = σ | B, σ0 : B → B′ = σ0(B) ⊆ E. Then�(σ) = [σ0], so � is onto.

(3) Clearly Gal(B/F) ⊆ QB as Gal(B/F) = {σ0 ∈ QB | B′ = B}.The first sentence then follows directly from Corollary 2.8.11. As for the sec-ond, if �(σ) = [σ0] and �(τ) = τ0, then σ | B = σ0 and τ | B = τ0,so στ | B = σ0τ0 and hence �(στ) = [σ0τ0]. But if GB is normal,(σ0GB)(τ0GB) = σ0τ0GB, i.e., [σ0τ0] = [σ0][σ0], so �(στ) = �(σ)�(τ).Thus � is a group homomorphism that is one-to-one and onto as a map ofsets, so � is an isomorphism. ��Proposition 2.8.14. Let E be a finite Galois extension of F and let B1 and B2

be intermediate fields with B1 = Fix(H1) and B2 = Fix(H2).Then:(1) B1B2 = Fix(H1 ∩ H2)

(2) B1 ∩B2 = Fix(H3) where H3 is the subgroup generated by H1 and H2.

Proof. (1) If σ ∈ H1 ∩ H2, then σ ∈ H1, so σ fixes B1, and σ ∈ H2, so σ fixesB2; hence σ fixes B1B2. On the other hand, if σ fixes B1B2, then σ fixes B1,so σ ∈ H1, and σ fixes B2, so σ ∈ H2; hence σ ∈ H1 ∩ H2.

(2) If σ(β) = β for every σ ∈ H3, then σ(β) = β for every σ ∈ H1, soβ ∈ B1, and σ(β) = β for every σ ∈ H2, so β ∈ B2; hence β ∈ B1 ∩ B2. Onthe other hand, if β ∈ B1 ∩ B2, then b ∈ B1, so σ(β) = β for every σ ∈ H1,and b ∈ B2, so σ(β) = β for every σ ∈ H2; hence σ(β) = β for everyσ ∈ H3. ��

In the original development of group theory, and in particular in the workof Galois, groups were not regarded as abstract groups, but rather as groupsof symmetries, and in particular as groups of permutations of the roots of apolynomial. This viewpoint, still a most useful one, is encapsulated in the fol-lowing result. In order to state it, we first recall that an action of a group G ona set S = {si } is transitive if for any two elements si and s j of S, there is anelement σ of G with σ(si ) = s j .

Proposition 2.8.15. Let E be a finite Galois extension of F, so E is the splittingfield of a separable polynomial f (X) ∈ F[X ]. Then G = Gal(E/F) is isomor-phic to a permutation group on the roots of f (X). If f (X) is irreducible, thenG is isomorphic to a transitive permutation group on the roots of f (X).


Proof. Let α1, . . . , αd be the roots of f (X) in E. For any σ ∈ Gal(E/F),σ(αi ) is a root of σ( f (X)) = f (X), so Gal(E/F) permutes {αi }. Furthermore,E = F(α1, . . . , αd) (Remark 2.5.6), so if σ(αi ) = αi for each i , then σ = id.

Now if f (X) is irreducible, and αi and α j are any two roots of f (X),there is isomorphism σ0 : F(αi ) → F(α j ) with σ | F = id and σ0(αi ) = α j

(Lemma 2.6.1), and σ0 extends to σ : E → E (Lemma 2.6.3). ��Remark 2.8.16. Note Proposition 2.8.15 implies Lemma 2.5.3 in case f (X) isseparable. �Remark 2.8.17. Note that Proposition 2.8.15 is about how the Galois group isrealized, not about its structure as an abstract group. For example, considerE = Q(

√2,

√3) and F = Q. Then G = Gal(E/F) ∼= (Z/2Z) ⊕ (Z/2Z),

and E is the splitting field of the polynomial (X2 − 2)(X2 − 3), which is notirreducible, so G does not act transitively on {√2, −√

2,√

3, −√3}. On the

other hand E = Q(√

2 + √3) and E is the splitting field of the polynomial

(X − (√

2 +√3))(X − (

√2 −√

3))(X − (−√2 +√

3))(X − (−√2 −√

3)) =X4 − 10X2 + 1, which is irreducible, and G does indeed act transitively on{√2 + √

3,√

2 − √3, −√

2 + √3, −√

2 − √3}. �

Corollary 2.8.18. Let f (X) ∈ F[X ] be a separable polynomial and writef (X) = f1(X)e1 · · · fk(X)ek , a factorization into irreducibles in F[X ]. Letfi (X) have degree di . Let E be the splitting field of f (X) ∈ F[X ]. ThenGal(E/F) is isomorphic to a subgroup of Sd1 × · · · × Sdk (where Sd denotesthe symmetric group on d elements). In particular, if f (X) ∈ F[X ] is a sepa-rable polynomial of degree d, and E is the splitting field of f (X) ∈ F[X ], thenGal(E/F) is isomorphic to a subgroup of Sd . ��Theorem 2.8.19. Let E be a finite Galois extension of F of degree d. ThenG = Gal(E/F) is isomorphic to a transitive subgroup of Sd , the symmetricgroup on d elements. Also, G has a subgroup H of index d.

Proof. Since E is a Galois extension of F, it is certainly a separable extensionof F (Proposition 2.8.14).We now apply a result we shall prove later (Corollary3.5.3) to conclude that E = F(α) for some α ∈ E. Then d = (E/F) =(F(α)/F) = deg mα(X) (Proposition 2.4.6 (2)), so the theorem follows fromProposition 2.8.15 and the following general fact: If G acts transitively onS = {si }, a set of cardinality d, then H = {σ ∈ G | σ(s1) = s1} is a subgroupof index d. ��

2.9 Examples 37

2.9 Examples

In this section we give a number of examples of the Fundamental Theorem ofGalois Theory, and compute a number of Galois groups.

Example 2.9.1. See Example 1.1.1. �Example 2.9.2. See Example 1.1.2. �Example 2.9.3. See Example 1.1.3. �Example 2.9.4. See Example 1.1.4. �Example 2.9.5. Let Ea be the splitting field of f (X) = X3 − a over Q, wherefor simplicity we take a to be an integer. There are several cases.

a = 0: Then f (X) = (X)(X)(X) is a product of linear factors, so Ea =Q. (Ea is obtained from Q by adjoining 0 three times, but adjoining 0 doesnothing.) Note (Ea/Q) = 1.

a �= 0 a perfect cube: Let a = b3. Then f (X) = X3−b3 = b3((X/b)3−1)

so Ea = E1. Thus we need only find the splitting field of f (X) = X3 −1. Thiswe have done already. E1 = Q(ω) where ω = (−1 + i

√3)/2 is a primitive

cube root of 1. Note (Ea/Q) = 2 and Gal(Ea/Q) ∼= Z/2Z.

a �= 0 not a perfect cube: This situation is entirely analogous to Exam-ple 1.1.3. Ea = Q(ω, 3

√a) and (Ea/Q) = 6. Also, G = Gal(Ea/Q) ∼= D6, the

dihedral group of order 6. Note that we can determine G without doing anycomputation. We know that |G| = (Ea/Q) = 6, and there are only two groupsof order 6, Z/6Z, which is abelian, and D6, which is not. Now Ea ⊃ Q( 3

√a)

and Q( 3√

a) is not a normal extension of Q, so G has a subgroup that is notnormal and hence G is nonabelian. �Example 2.9.6. E is the splitting field of f (X) = X6 − 3 over Q. Then E =Q(ζ,

6√

3) where ζ is a primitive sixth root of 1 and (E/Q) = 12. We maychoose ζ = (1 + i

√3)/2. Then ω = ζ 2 and −1 = ζ 3. But then notice also

that ζ = 1 + ω so in particular Q(ζ ) = Q(ω). An element σ of Gal(E/Q) isdetermined by its effect on ζ and on 6

√3. One can check that G = Gal(E/Q)

has elements σ and τ with:

σ(ζ ) = ζ−1 τ(ζ ) = ζ

σ (6√

3) = 6√

3 τ(6√

3) = ζ6√

3.

Clearly σ 2 = 1 and τ 6 = 1. Also στ(ζ ) = ζ 5 = τ−1σ(ζ ) and στ(6√

3) =ζ−1 6

√3 = τ−1σ(

6√

3). Since any element of G must take ζ to ζ or ζ−1, and


must take 3√

6 to ζ i 3√

6 for some i , we see that σ and τ generate G. Thus weconclude that

G = 〈σ, τ | σ 2 = 1, τ 6 = 1, i στ = τ−1σ 〉is isomorphic to the dihedral group D12. It is then routine but lengthy to checkthat G has 15 subgroups, which we list below, where subgroups denoted by thesome letter are mutually conjugate (so in particular A1, D1, E1, G1, H1 and I1

are normal):A1 = {1}B1 = {1, σ }, B2 = {1, τ 2σ }, B3 = {1, τ 4σ }C1 = {1, τσ }, C2 = {1, τ 3σ }, C3 = {1, τ 5σ }D1 = {1, τ 3}E1 = {1, τ 2, τ 4}F1 = 1, σ, τ 3, τ 3σ }, F2 = {1, τσ, τ 3, τ 4σ }, F3 = {1, τ 2σ, τ 3, τ 5σ }G1 = {1, τ 2, τ 4, σ, τ 2σ, τ 4σ }H1 = {1, τ, τ 2, τ 3, τ 4, τ 5}I1 = G

These subgroups have the following fixed fields:Fix(A1) = EFix(B1) = Q(

6√

3), Fix(B2) = Q(ζ6√

3), Fix(B3) = Q(ω6√

3)

Fix(C1) = Q((1 + ζ )6√

3), Fix(C2) = Q(i 6√

3),Fix(C3) = Q((1 + ζ−1)

6√

3)

Fix(D1) = Q(ζ,3√

3)

Fix(E1) = Q(ζ,√

3)

Fix(F1) = Q(3√

3), Fix(F2) = Q(ζ3√

3), Fix(F3) = Q(ω3√

3)

Fix(G1) = Q(√

3)

Fix(H1) = Q(ζ )

Fix(I1) = Q

The 6 subfields fixed by normal subgroups are Galois extensions of Q, so aresplitting fields of separable polynomials:

Fix(A1) = splitting field of X6 − 3Fix(D1) = splitting field of X3 − 3Fix(E1) = splitting field of (X2 − X + 1)(X2 − 3)

Fix(G1) = splitting field of X2 − 3Fix(H1) = splitting field of X2 − X + 1Fix(I1) = splitting field of 1

Note here there are also interesting relationships between the fields interme-diate between E and Q. For example, let F = Fix(F1) = Q(

3√

3). Then B1,

2.9 Examples 39

C2, and D1 are all normal subgroups of F1 (although B1 and C2 are not nor-mal subgroups of G). Hence Fix(B1)/F1, Fix(C2)/F1, and Fix(D1)/F1 are allGalois extensions, all of degree two, hence with Galois groups isomorphic toZ/2Z, and they are the splitting fields of the polynomials X2 − 3

√3, X2 + 3

√3,

and X2 + X + 1 ∈ Q(3√

3)[X ], respectively. �Example 2.9.7. E is the splitting field of X6 + 3 over Q. Again E =Q(ζ,

6√−3) where ζ is a primitive sixth root of 1. However, (

6√−3)3 = i

√3,

so ζ ∈ Q(6√−3) and E = Q(

6√−3). Thus in this case (E/Q) = 6, and here

Gal(E/Q) = D6 by an argument as in Example 2.9.5.Let us examine this example more closely. Let α = 6

√−3. Then E = Q(α),so λ ∈ G = Gal(E/Q) is determined by λ(α). Since λ(α) must be a root ofX6 + 3, we must have λ(α) = ζ kα, for some k, where ζ = (1 + i

√3)/2 is

a primitive sixth root of 1, and all possible values of k, 0 ≤ k ≤ 5 occur. Ifλ(α) = ζ kα, with 0 ≤ k ≤ 5, set λ = λk . Thus G = {id, λ1, . . . , λ5}. Thismakes G “look” abelian, but we can see by direct calculation that it is not.

Note that ζ = (1 + α3)/2, so

λ j (ζ ) = (1 + λ j (α3))/2 = (1 + (λ j (α))3)/2 = (1 + (ζ jα)3)/2

= (1 + ζ 3 jα3)/2 = ζ for j even, ζ−1 for j odd.

In particular we see that

λ jλi (α) = λ j (λi (α)) = λ j (ζiα) = λ j (ζ

i )λ j (α) = λ j (ζ )iζ jα

= ζ kα where k = (−1) j i + j,

directly exhibiting the noncommutativity of G.In fact, we may obtain more precise information on the structure of G,

again by direct computation. Note that ζ 2 = ω = (−1 + i√

3)/2, so, settingγ = i

√3 = √−3, we see that Q(ζ ) = Q(ω) = Q(γ ). Note also that α =

γ /β. Then E = Q(α) = (Q(β))(γ ) = (Q(γ ))(β). Hence there is an elementϕ of G with ϕ(γ ) = −γ, ϕ(β) = β, and an element ψ of G with ψ(γ ) =γ, ψ(β) = ωβ. Then

ϕ(α) = ϕ(γ /β) = ϕ(γ )/ϕ(β) = −γ /β = ζ 3(γ /β) = ζ 3α so ϕ = λ3,

ψ(α) = ψ(γ/β) = ψ(γ )/ψ(β) = γ /(ωβ) = ζ 4(γ /β) = ζ 4α so ψ = λ4.

Observe that ϕ2 = id, ψ2 = λ2, and ψ3 = id. Furthermore

ϕ−1ψϕ(α) = ϕ−1ψϕ(γ /β) = ϕ−1ψ(−γ /β)

= ϕ−1(−γ /(ωβ)) = γ /(ω2β) = ωα = ζ 2α,


so we conclude λ−13 λ4λ3 = λ2

4, giving an explicit presentation of G as thedihedral group D6. �Example 2.9.8. If E is the splitting field of an irreducible cubic polynomialover Q, then, by Theorem 2.8.19, Gal(E/Q) is a subgroup of S3 whose or-der is divisible by 3, so it is either S3 or Z/3Z. The construction in Exam-ple 2.9.5 always gave S3. Here is a case where the Galois group is Z/3Z.Let ζ be a primitive ninth root of 1 and consider the polynomial f (X) =(X − (ζ + ζ−1))(X − (ζ 2 + ζ−2))(X − (ζ 4 + ζ−4)) = X3 − 3X + 1 ∈ Q[X ].This polynomial is irreducible over Q (as otherwise it would have a linearfactor, i.e., a root in Q, and it does not). Let λ0, λ1, and λ2 be the roots ofthis polynomial, in the order written, and let E = F(λ0, λ1, λ2) be its splittingfield.

Then (E/Q) = deg f (X) = 3, and since | Gal(E/Q)| = (E/Q), the Ga-lois group must be isomorphic to Z/3Z. We can see this concretely as follows:Note that λ2

0 = λ1 + 2 and λ21 = λ2 + 2 (and also λ2

2 = λ0 + 2), so in fact E =F(λ0), and σ ∈ Gal(E/Q) is determined by σ(λ0). Since f (X) is irreducible,there is an element σi ∈ Gal(E/Q) with σi (λ0) = λi , i = 0, 1, 2. Clearly σ0 =id. If σ1(λ0) = λ1, then σ1(λ1) = σ1(λ

20 −2) = σ1(λ0)

2 −σ1(2) = λ21 −2 = λ2

and similarly σ1(λ2) = λ0. Then σ 21 (λ0) = σ1(σ1(λ0)) = σ1(λ1) = λ2, so

σ 21 = σ2; similarly σ 3

1 (λ0) = λ0, so σ 31 = σ0, and we see explicitly that

Gal(E/Q) = {1, σ1, σ 21 }. �

2.10 Exercises

Exercise 2.10.1. In each case, find a polynomial in Q[X ] that has α as a root.Then find all complex roots of this polynomial.

(a) α = 1 + √2.

(b) α = √2 + √

3.

(c) α =√

1 + √2.

(d) α =√

3 − 2√

2.

(e) α = 3√

1 + √−3.

(f) α = 3√

1 + √2.

(g) α =√

1 + 3√

2.

(h) α =√

6 + √3 +

√6 − √

3.

(i) α =√

4 + √7 +

√4 − √

7.

2.10 Exercises 41

Exercise 2.10.2. Let f (X) = X3 + 7X + 1 ∈ Q[X ]. This polynomial is ir-reducible. Let E = Q[α] for some root α of f (X). Consider the followingelements of E: β1 = α + 3, β2 = α2 + 1, β3 = α2 − 2α + 3.(a) Express each of the following in the form g(α) for some polynomialg(X) ∈ Q[X ]: β−1

1 , β−12 , β−1

3 , β1β2, β1β3, β2β3.(b) Find mβ1(X), mβ2(X), mβ3(X) ∈ Q[X ].Exercise 2.10.3. Let f (X) = X3 + X2 + 2 ∈ F5[X ]. This polynomial is ir-reducible. Let E = F5[α] for some root α of f (X). Consider the followingelements of E: β1 = 2α + 1, β2 = α2 − 1, β3 = α2 − α + 2.(a) Express each of the following in the form g(α) for some polynomialg(X) ∈ F5[X ]: β−1

1 , β−12 , β−1

3 , β1β2, β1β3, β2β3.(b) Find mβ1(X), mβ2(X), mβ3(X) ∈ F5[X ].Exercise 2.10.4. As we shall see in Section 3.3, for every prime power q = pr ,there is a unique (up to isomorphism) field Fq with q elements.(a) Write down an explicit addition and multiplication table for F2, F3, and F5.(b) Write down an explicit addition and multiplication table for F4 and F9.(c) For each element α ∈ F4, find its minimum polynomial mα(X) ∈ F2[X ],and for each element α ∈ F9, find its minimum polynomial mα(X) ∈ F3[X ].(d) Explicitly compute the Frobenius endomorphism on F4 and F9.(e) Write down an explicit addition and multiplication table for F8.(f) For each element α ∈ F8, find its minimum polynomial mα(X) ∈ F2[X ].(g) Explicitly compute the Frobenius endomorphism on F8.

Exercise 2.10.5. (a) Find all irreducible quadratic, cubic, and quartic polyno-mials f (X) ∈ F2[X ].(b) Find all irreducible quadratic, and cubic polynomials f (X) ∈ F3[X ].(c) Find all irreducible quadratic polynomials f (X) ∈ F5[X ].(d) Find all irreducible quadratic polynomials f (X) ∈ F4[X ].Exercise 2.10.6. Let f (X) = an Xn + · · · + a0 ∈ F[X ] split in E. Assume thatchar(F) �= 2. Show that (1) implies (2):(1) For every α ∈ E, if α is a root of f (X), then −α ∈ E is also a root of f (X)

of the same multiplicity.(2) an−k = 0 for all odd k.

Exercise 2.10.7. Let f (X) = an Xn + · · · + a0 ∈ F[X ] split in E. Assume thata0 �= 0. Show that (1) implies (2):(1) For every α �= 0 ∈ E, if α is a root of f (X), then 1/α ∈ E is also a root off (X) of the same multiplicity.(2) For e = ±1, an−k = eak for all k.


Exercise 2.10.8. (a) Let f (X) ∈ Z[X ] be a monic polynomial. Show that ifits mod p reduction f (X) ∈ Fp[X ] is irreducible for some p, then f (X) isirreducible.(b) Use this to show that f (X) = X3 − 5X + 36 is irreducible.

Exercise 2.10.9. A priori, there is a distinction between polynomial expres-sions and polynomial functions, i.e., between f (X) regarded as an element ofF[X ] and f (X) as defining a function on F. Clearly, if f (X) = g(X) as ele-ments of F[X ], then f (X) = g(X) as functions on F[X ] (i.e., f (a) = g(a)

for every a ∈ F).(a) Show that, if F is infinite, the converse is true: If f (a) = g(a) for everya ∈ F, then f (X) = g(X) ∈ F[X ].(b) Find a counterexample to the converse if F is finite. That is, if F is finite,find polynomials f (X) �= g(X) ∈ F[X ] but with f (a) = g(a) for everya ∈ F.(c) Use the result of (b) to show that no finite field is algebraically closed, i.e.,that if F is any finite field, there is a nonconstant polynomial f (X) ∈ F[X ]that does not have a root in F.

Exercise 2.10.10. (a) Let f (X) ∈ F[X ] have roots α1, . . . , αn in some split-ting field E. Show that f (X) is irreducible in F[X ] if and only if∏

αi ∈T

(X − αi ) /∈ F[X ]

for any nonempty proper subset T of S = {α1, . . . , αn}.(b) Observe that E = Q[√3,

√5] is the splitting field of f1(X) = X4−46X2+

289, with roots√

3 + 2√

5,√

3 − 2√

5, −√3 + 2

√5, and −√

3 − 2√

5, andalso of f2(X) = X4 − 6X3 + 5X2 + 10X + 2, with roots 1 + √

3, 1 − √3,

2 + √5, and 2 − √

5. Use part (a) to show that f1(X) is irreducible in Q[X ]but f2(X) is not.

Exercise 2.10.11. Let f (X), g(X) ∈ F[X ] be monic irreducible polynomials.Show that if f (X) and g(X) have a common root in some extension E of F,then f (X) = g(X).

Exercise 2.10.12. Let E ⊇ F and let α �= 0 ∈ E. Show that α is algebraic overF if and only if 1/α = f (α) for some polynomial f (X) ∈ F[X ].Exercise 2.10.13. Let R be a subring of F and let E be an extension of F. Letα ∈ E be algebraic over F. If mα(X) ∈ R[X ], then α is called integral overR. In case R = Z and F = Q, show that α ∈ E is integral over R if andonly if f (α) = 0 for some monic polynomial f (X) ∈ Z[X ]. (Such a complexnumber is called an algebraic integer.)

2.10 Exercises 43

Exercise 2.10.14. Suppose that R = Z and F = Q. Show that, for any α ∈ E,there is an integer n �= 0 ∈ Z such that nα is an algebraic integer.

Exercise 2.10.15. Let E be a field and let R be a subring of E. If every elementof E is integral over R, show that R is a field.

Exercise 2.10.16. Let E be an extension of F and let f (X) ∈ F[X ] be an ir-reducible polynomial. Let f (X) = f1(X) f2(X) · · · ft (X) ∈ E[X ] be factoredas a product of irreducible polynomials.(a) If E is a Galois extension of F, show that { f1(X), f2(X), . . . , ft (X)} aremutually conjugate in E[X ]. (Two polynomials f1(X), f2(X) ∈ E[X ] are mu-tually conjugate if there exists an element σ ∈ Gal(E/F) with σ( f1(X)) =f2(X).)(b) Give a counterexample to this if E is not a Galois extension of F.

Exercise 2.10.17. Let E be a splitting field of the separable irreducible poly-nomial f (X) ∈ F[X ].(a) For a root α ∈ E of f (X), let r(α) be the number of roots of f (X) in F(α).Show that r(α) is independent of the choice of α. Call this common value r .(b) Let d be the number of distinct fields {F(α) | α ∈ E a root of f (X)}. Showthat dr = deg f (X).(c) Give examples where r = 1, 1 < r < deg f (X), and r = deg f (X).

Exercise 2.10.18. Let E be the splitting field of the separable irreducible poly-nomial f (X) ∈ F[X ] and let B be a Galois extension of F with F ⊆ B ⊆ E.Let f (X) = f1(X) · · · fk(X) ∈ B[X ] be a factorization into irreducible poly-nomials. Show that each fi (X) has the same degree d. Furthermore, if α ∈ Eis any root of f (X), show that d = (B(α)/B) and that k = (B ∩ F(α)/F).

Exercise 2.10.19. Let f (X) ∈ Q[X ] be an irreducible cubic with roots α, β, γ

in some splitting field E. Show that one of the following two alternatives holds:(1) Gal(E/Q) ∼= Z/3Z, and there is no field B with Q ⊂ B ⊂ E.(2) Gal(E/Q) ∼= S3, and there are exactly four fields B with Q ⊂ B ⊂ E,these fields being Q(α), Q(β), Q(γ ), and Q(

√D) for some D ∈ Q that is not

a perfect square.(In fact, it is possible to decide between these two alternatives, and to find D,without finding the roots of f (X). See Section 4.8.)

Exercise 2.10.20. Let f (X) = Xn − p ∈ Q[X ]. By Proposition 4.1.7 below,f (X) is irreducible. Let E be a splitting field of f (X). Show that for everyn ≥ 3, the Galois group G = Gal(E/Q) is not abelian.


Exercise 2.10.21. Let E be a splitting field of the separable irreducible polyno-mial f (X) ∈ F[X ]. Let α be any root of f (X) in E. Let p be a prime dividing(E/F). Show there is a field F ⊆ B ⊆ E with E = B(α) and with (E/B) = p.

Exercise 2.10.22. Let F be an arbitrary field. Let m and n be positive integersand let d = gcd(m, n). Let f (X) = Xm − 1, g(X) = Xn − 1 ∈ F[X ]. Showthat gcd( f (X), g(X)) = h(X) where h(X) = Xd − 1 ∈ F[X ].Exercise 2.10.23. Let f (X) ∈ Q[X ]. If f (X) ∈ Z[X ], then certainlyf (n) ∈ Z for every n ∈ Z. The converse of this is false, as the examplef (X) = X (X + 1)/2 shows. Let c0(X) = 1 and let ck(X) = X (X − 1) · · ·(X −(k−1))/k! for k > 0. Show that any f (X) ∈ Q[X ] with the property thatf (n) ∈ Z for every n ∈ Z can be written a linear combination of {ck(X)} withinteger coefficients. (This remains true if Q is replaced by any field of charac-teristic 0.) Note this implies that if f (n) is a polynomial of degree at most kwith f (n) an integer divisible by k! for every integer n, then f (X) ∈ Z[X ].

3

Development and Applications of Galois Theory

3.1 Symmetric Functions and the Symmetric Group

We now apply our general theory to the case of symmetric functions. We letD be an arbitrary field and set E = D(X1, . . . , Xd), the field of rational func-tions in the variables X1, . . . , Xd . Then the symmetric group Sd acts on E bypermuting {X1, . . . , Xd}.Definition 3.1.1. The subfield F of E fixed under the action of Sd is thefield of symmetric functions in X1, . . . , Xd . (F = { f (X1, . . . , Xd) ∈ E |f (Xσ(1), . . . , Xσ(d)) = f (X1, . . . , Xd) for every σ ∈ Sd}.) A polynomialf (X1, . . . , Xd) ∈ F is a symmetric polynomial. �

Since F = Fix(Sd), by definition E is a Galois extension of F with Galoisgroup Sd . We may use this observation to easily show that any finite group isthe Galois group of some field extension.

Lemma 3.1.2. (1) Let G be a finite group. Then there is a Galois extensionE/B with Gal(E/B) ∼= G.

(2)There is a Galois extension E/B with Gal(E/B) ∼= G and with E (andhence B) a subfield of the complex numbers C.

Proof. (1) Any finite group G is isomorphic to a subgroup of some sym-metric group Sd . Identify G with its image under this isomorphism. LetE = D(X1, . . . , Xd), as above, and set B = Fix(G). Then E/B is Galoiswith Galois group isomorphic to G.

(2) Let D = Q. Then we need only show there is subfield of Cisomorphic to Q(X1, . . . , Xd). Let t1, . . . , td be algebraically independentelements of C, i.e., elements that do not satisfy any polynomial equation.

S.H. Weintraub, Galois Theory, DOI 10.1007/978-0-387- _3, © Springer Science+Business Media, LLC 2009

87575-0

46 3 Development and Applications of Galois Theory

Let ϕ : Q(X1, . . . , Xd) → Q(t1, . . . , td) be defined by ϕ(Xi ) = ti . Ifr(X1, . . . , Xd) = f (X1, . . . , Xd)/g(X1, . . . , Xd) ∈ Q(X1, . . . , Xd), thenϕ(r(X1, . . . , Xd)) = f (t1, . . . , td)/g(t1, . . . , td) = r(t1, . . . , td). Then ϕ iswell defined as the denominator g(t1, . . . , td) is never zero. It is obviouslyonto and it is 1–1 as the numerator f (t1, . . . , td) is never zero, so ϕ is an iso-morphism. Then set E = Q(t1, . . . , td) and let B be the fixed field of G actingby permuting t1, . . . , td .

It remains only to show that we can indeed find algebraically indepen-dent elements t1, . . . , td of Q. Let E0 = Q and consider the extension fieldF1 = {z ∈ C | z is algebraic over E0}. E0 is countable; so there are only count-ably many polynomials in E0[X1], each of which has only finitely many roots.Hence F1 is countable. But C is uncountable. Thus we may choose any t1 /∈ F1

and let E1 = E0(t1). Then X1 �→ t1 induces an isomorphism E0(X1) → E1.But again E0(X1) has only countably many elements, so E1 is countable andwe may iterate this process. ��

We now return to the study of symmetric functions.

Definition 3.1.3. The elementary symmetric polynomials s1, . . . , sd are

sk =∑

1≤i1<...<ik≤d

Xi1 Xi2 · · · Xik , k = 1, . . . , d. �

For example, if d = 4:

s1 = X1 + X2 + X3 + X4,

s2 = X1 X2 + X1 X3 + X1 X4 + X2 X3 + X2 X4 + X3 X4,

s3 = X1 X2 X3 + X1 X2 X4 + X1 X3 X4 + X2 X3 X4,

s4 = X1 X2 X3 X4.

Lemma 3.1.4. The field of symmetric functions F = D(s1, . . . , sd).

Proof. Let B = D(s1, . . . , sd). Clearly F ⊇ B, as each si is fixed by Sd .Now (E/F) = |Sd | = d!, so (E/B) ≥ d!. We will show that F is the

splitting field of a separable polynomial f (X) ∈ B[X ] of degree d. Then, byLemma 2.8.18, (E/B) = | Gal(E/B)| ≤ d!. Hence (E/B) = d! and F = B.

Now f (X) is simply the polynomial

f (X) = (X − X1)(X − X2) · · · (X − Xd),

which obviously has splitting field E, as its roots are X1, . . . , Xd . ��

3.1 Symmetric Functions and the Symmetric Group 47

Remark 3.1.5. (1) Note in particular that we have constructed an explicit poly-nomial f (X) of degree d whose Galois group is Sd . Explicit computationshows that

f (X) = Xd +d∑

i=1

(−1)i si Xd−i .

(2) Let us write f (X) = Xd + ad−1 Xd−1 +· · ·+ a0. Then we see from theabove expression that the coefficients of f (X) are, up to sign, the elementarysymmetric functions of its roots,

ad−i = (−1)i si , i = 1, . . . , d,

and, in fact, this is true for any polynomial f (X). �Lemma 3.1.6. Let Y be the monomial

Y = X12 X2

3 · · · Xd−1d (= X0

1 X12 · · · Xd−1

d ).

Then E = F(Y ).

Proof. Obviously, the only element of Sd that fixes Y is the identity, so Yhas d! conjugates. Then, by Corollary 2.7.13, (F(Y )/F) = d! = (E/F), soE = F(Y ). ��Corollary 3.1.7. Any element f (X1, . . . , Xd) of E can be written uniquely as

f (X1, . . . , Xd) =d!−1∑i=0

gi (si , . . . , sd)Yi

where gi (s1, . . . , sd) is a rational function in s1, . . . , sd .

Proof. By Lemma 3.1.6 and Proposition 2.4.6 (1), {1, Y, . . . , Y d!−1} is a vec-tor space basis for E over F, and any element of F is a rational function ofs1, . . . , sd , by Lemma 3.1.4. ��Remark 3.1.8. In the language of Section 3.5, Lemma 3.1.6 shows that E is asimple extension of F and that Y is a primitive element of E. �Lemma 3.1.9. Let Y be the monomial Y = X1

2 · · · Xd−1d (as in Lemma 3.1.6)

and for σ ∈ Sd , let Yσ be the monomial

Yσ = X1σ(2) · · · Xd−1

σ(d).

Then {Yσ | σ ∈ Sd} is a (vector space) basis for E over F.


Proof. We prove the claim by induction on d.For d = 1 this is simply the claim that Y = 1 is a basis for E over E, which

is trivial.Now assume the result is true for d − 1, and suppose that

∑σ∈Sd

cσ (X1, . . . , Xd)Xσ(2) · · · Xd−1σ(d) = 0,

with cσ (X1, . . . , Xd) ∈ F. We wish to show that cσ (X1, . . . , Xd) = 0for each σ . Assume not. By “clearing denominators” we may assume that{cσ (X1, . . . , Xd)} are polynomials in F with no common polynomial factor.

For convenience, write the above sum as∑

. Let Hi = {σ ∈ Sd | σ(i) =1}, i = 1, . . . , d . Then

∑ = ∑1 + · · · + ∑

d , where∑

i

=∑σ∈Hi

cσ (X1, . . . , Xd)Xσ(2) · · · Xd−1σ(d).

We may further write∑

i = ∑′i + ∑′′

i , i = 1, . . . , d, where∑′

i istaken over those σ ∈ Hi with cσ (X1, . . . , Xd) not divisible by Xi , and

∑′′i

is taken over those σ ∈ Hi with cσ (X1, . . . , Xd) divisible by Xi . Call thesetwo subsets H ′

i and H ′′i , respectively.

Let us consider the equation∑ = 0. Note that the exponent of X1 is at

least one for each term in∑

i , i �= 1 (as if σ(i) �= 1, then σ( j) = 1 for somej with 2 ≤ j ≤ d), and also for each term in

∑′′1. Thus, considering the terms

in∑

that are not divisible by X1, we thus see that we must have∑′

1 = 0, i.e.,that

∑σ∈H ′

1

cσ (X1, . . . , Xd)Xσ(2) · · · Xd−1σ(d) = 0.

Observe that H1 is isomorphic to Sd−1, permuting the subset {2, . . . , d} of{1, . . . , d} (and each Hi is a left coset of H1).

Let us set X1 = 0 in∑′

1, thereby obtaining

0 =∑σ∈H ′

1

cσ (0, X2, . . . , Xd)Xσ(2) . . . Xd−1σ(d)

=( ∑

σ∈H ′1

cσ (0, X2, . . . , Xd)Xσ(3) · · · Xd−2σ(d)

)X2 · · · Xd .

Now we apply the inductive hypothesis. {Xσ(3) · · · Xd−2σ(d)} is a basis for

D(X2, . . . , Xd) over the subfield of symmetric functions in X2, . . . , Xd , so inparticular they are linearly independent. Hence we have that each coefficient


cσ (0, X2, . . . , Xd) = 0. In other words, each cσ (X1, . . . , Xd) is a polynomialthat vanishes when we set X1 = 0, so it must have X1 as a factor, contra-dicting the definition of H ′

1. Hence we see that H ′1 = ∅, and so H ′′

1 = H1.In other words, cσ (X1, . . . , Xd) is divisible by X1 for every σ ∈ H1. Butcσ (X1, . . . , Xd) ∈ F is a symmetric polynomial in X1, . . . , Xd , so this impliesthat cσ (X1, . . . , Xd) is also divisible by X2, . . . , Xd , and hence by the productX1 · · · Xd , for each σ ∈ H1.

Similarly, we see, for each i = 1, . . . , d, that H ′i = ∅, H ′′

i = Hi ,so cσ (X1, . . . , Xd) is divisible by Xi , and thus by X1 · · · Xd , for eachσ ∈ Hi . Thus {cσ (X1, . . . , Xd) | σ ∈ Sd} have the common polyno-mial factor X1 · · · Xd , contradicting our choice of {cσ (X1, . . . , Xd)}. Thus,{Xσ(2) · · · Xd−1

σ(d)} is linearly independent, and thus is a vector space basis forE over F (as it has d! = (E/F) elements), completing the inductive step. ��Corollary 3.1.10. Any element f (X1, . . . , Xd) of E can be written uniquelyas

f (X1, . . . , Xd) =∑σ∈Sd

hσ (s1, . . . , sd)Xσ(2) · · · Xd−1σ(d)

where hσ (s1, . . . , sd) is a rational function of s1, . . . , sd .

Proof. Lemma 3.1.4 and Lemma 3.1.9. ��Remark 3.1.11. In the language of Section 3.6, Lemma 3.1.9 shows that{Xσ(2) · · · Xd−1

σ(d) | σ ∈ Sd} is a normal basis of E over F. �Our discussion here is not yet complete. Corollaries 3.1.7 and 3.1.10 show

how to write rational functions uniquely in terms of symmetric functions. Whatwe have not yet shown is that symmetric functions can be written uniquely interms of the elementary symmetric functions. We do that now.

Lemma 3.1.12. Let f (X1, . . . , Xd) ∈ F be a symmetric function. Thenf (X1, . . . , Xd) may be written uniquely as a rational function e(s1, . . . , sd)

of the elementary symmetric functions s1, . . . , sd .

Proof. We prove this by induction on d. In case d = 1, s1 = X1 and e(s1) =f (X1) is the unique expression. Now suppose the result is true for d − 1.

We first need to make an observation about the elementary symmetric func-tions. We really should have written sk as sk,d , as the expressions in Defini-tion 3.1.4 depended on d , but for simplicity we did not. We do that now. Wethen observe that if we let sk,d be the expression obtained from sk,d by settingXd = 0, then


sk,d = sk,d−1 for k ≤ d − 1 and sd,d = 0.

By equating different expressions for f (X1, . . . , Xd) and “clearing de-nominators”, it is easy to see that to prove the lemma it suffices to showthat the only polynomial e(s1,d , . . . , sd,d) that is identically zero is the zeropolynomial. Thus suppose e(s1,d , . . . , sd,d) = 0 but e(s1,d , . . . , sd,d) is notthe zero polynomial. Since F is a field and sd,d is a nonzero element of F,e(s1,d , . . . , sd,d) = 0 if and only if (sd,d)

−ke(s1,d , . . . , sd,d) = 0 for any k.Hence, factoring out by a power of sd,d if necessary, we may assume, expand-ing e(s1,d , . . . , sd,d) in powers of sd,d , that the constant term is nonzero. Inother words, we have that

0 = e(s1,d , . . . , sd,d) =n∑

i=0

ei (s1,d , . . . , sd−1,d)sid,d

with ei (s1,d , . . . , sd−1,d) polynomials in s1,d , . . . , sd−1,d , and furthermoree0(s1,d , . . . , sd−1,d) not the zero polynomial. Now set Xd = 0. Then the aboveexpressions become

0 = e(s1,d , . . . , sd−1,d , 0)

= e0(s1,d , . . . , sd−1,d)

= e0(s1,d−1, . . . , sd−1,d−1),

contradicting the d − 1 case. ��Remark 3.1.13. We have just shown that the only polynomial in s1, . . . , sd

that vanishes is the zero polynomial. This is usually phrased as saying thats1, . . . , sd are “algebraically independent” over D. �

We can also strengthen Lemma 3.1.12 as follows.

Lemma 3.1.14. Let f (X1, . . . , Xd) ∈ F be a symmetric polynomial. Thenf (X1, . . . , Xd) may be written uniquely as a polynomial e(s1, . . . , sd) in theelementary symmetric functions s1, . . . , sd .

Proof. Uniqueness follows from Lemma 3.1.12 so we must show existence.We do so by induction on d. We include in our inductive hypothesis that ev-ery term in e(s1, . . . , sd), when regarded as a polynomial in X1, . . . , Xd , hasdegree at most the degree of f (X1, . . . , Xd).

For d = 1, s1 = X1, and the result is trivial.Assume the result is true for d − 1. We proceed by induction on the degree

k of the symmetric polynomial. If k = 0 the result is trivial. Assume it is true


for all symmetric polynomials in X1, . . . , Xd of degree less than k, and letf (X1, . . . , Xd) have degree k.

Now f (X1, . . . , Xd−1, 0) is a symmetric polynomial in X1, . . . , Xd−1 soby induction f (X1, . . . , Xd−1) = g(s1,d−1, . . . , sd−1,d−1) for some polyno-mial g (where we use the notation of the proof of Lemma 3.1.12).

Then f (X1, . . . , Xd) − g(s1, . . . , sd−1) = f1(X1, . . . , Xd) is a sym-metric polynomial in X1, . . . , Xd with f1(X1, . . . , Xd−1, 0) = 0. Hencef1(X1, . . . , Xd) is divisible by Xd in D[X1, . . . , Xd ], and hence, since thispolynomial is symmetric, it is divisible by the product X1 . . . Xd as well.Hence f1(X1, . . . , Xd) = (X1 · · · Xd) f2(X1, . . . , Xd) with f2(X1, . . . , Xd)

a polynomial of lower degree. By induction f2(X1, . . . , Xd) = h(s1, . . . , sd)

for some symmetric polynomial h(s1, . . . , sd) satisfying our inductive hypoth-esis, and of course X1 · · · Xd = sd is a symmetric polynomial satisfying ourhypothesis as well.

Then

f (X1, . . . , Xd) = g(s1, . . . , sd−1) + sd h(s1, . . . , sd)

= e(s1, . . . , sd)

is a polynomial in s1, . . . , sd , as required. ��We have been considering polynomials with coefficients in a field. But, in

fact, we also have a result valid for polynomials with integer coefficients.

Lemma 3.1.15. Let f (X1, . . . , Xd) be a symmetric polynomial with integercoefficients. Then f (X1, . . . , Xd) may be written uniquely as e(s1, . . . , sd)

where e(s1, . . . , sd) is a polynomial in the elementary symmetric functionss1, . . . , sd with integer coefficients.

Proof. The proof of Lemma 3.1.14 goes through unchanged to prove thisresult. ��

In addition to the elementary symmetric functions, there is another set ofsymmetric functions that readily (perhaps even more readily) appears, namelysums of powers. The relationship between these is classical.

Theorem 3.1.16 (Newton’s Identities). Let {si }i=1,...,d be the elementarysymmetric polynomials in X1, . . . , Xd. Let ti = Xi

1 + · · · + Xid for i > 0.

Then:


t1 − s1 = 0

t2 − s1t1 + 2s2 = 0

t3 − s1t2 + s2t1 − 3s3 = 0

...

td − s1td−1 + · · · + (−1)d−1sd−1t1 + (−1)ddsd = 0

and

td+1 − s1td + · · · + (−1)dsd t1 = 0

td+2 − s1td+1 + · · · + (−1)dsd t2 = 0

...

Proof. Throughout this proof, we shall abbreviate∑d

j=1 by∑

.Let us set f (X) = (X − X1)(X − X2) · · · (X − Xd) = Xd + ad−1 Xd−1 +

· · · + a0. We know from Remark 3.1.5 that ad−i = (−1)i si , i = 1, . . . , d.From the division algorithm for polynomials we know that for any Y ∈ E,f (X) = (X − Y )q(X) + f (Y ) for a unique polynomial q(X) ∈ E of degreed −1. Let q(X) = bd−1 Xd−1 +bd−2 Xd−2 +· · ·+b0. Then long division showsthat

bd−1 = 1

bd−2 = Y + ad−1

bd−3 = Y 2 + ad−1Y + ad−2

...

(Of course, each bk is a function of Y , but we write bk instead of bk(Y ) forsimplicity. However, we need to keep this dependence in mind below.)

We now successively set Y = X1, . . . , Xd and form the sum

(∗)∑

q(X j ) =(∑

bd−1

)Xd−1 +

(∑bd−2

)Xd−2 + · · · .

On the one hand, we see from f (X) = (X − X1)(X − X2) · · · (X − Xd)

that∑

q(X j ) = f ′(X)

= d Xd−1 + (d − 1)ad−1 Xd−2 + (d − 2)ad−2 Xd−3 + · · · .


On the other hand, adding the expressions on the right-hand side of (∗),we obtain

∑q(X j ) = d Xd−1 +

(∑(X j + ad−1)

)Xd−2

+(∑

(X2j + ad−1 X j + ad−2)

)Xd−3 + · · · .

Comparing coefficients, we see

t1 + dad−1 = (d − 1)ad−1

t2 + ad−1t1 + dad−2 = (d − 2)ad−2

t3 + ad−1t2 + ad−2t1 + dad−3 = (d − 3)ad−3

which yields

t1 − s1 = 0

t2 − s1t1 + 2s2 = 0

t3 − s1t2 + s2t1 − 3s3 = 0

...

which are the first d − 1 Newton’s Identities. The remaining Newton’s Identi-ties comes from considering coefficients in

0 =∑

Xm f (X j )

for m = 0, 1, 2, . . . . ��Corollary 3.1.17. Let {si }, {ti }, and d be as in Theorem 3.1.16.

(1) For every k > 0, tk is a polynomial in s1, . . . , s j with integer coeffi-cients, where j = min(k, d).

(2) Let f (X1, . . . , Xd) ∈ F be a symmetric polynomial of degree at mostk, and let j = min(k, d). If char(D) = 0 or char(D) > j , then f (X) may bewritten uniquely as a polynomial in t1, . . . , tk .

(3) In particular, if char(D) = 0 or char(D) > d, then every symmetricpolynomial f (X1, . . . , Xd) ∈ F may be written uniquely as a polynomial int1, . . . , td .

Proof. Newton’s identities always give a “triangular” system for {ti }i=1,...,d interms of {si }i=1,...,d . They give a “triangular” system for {si }i=1,...,d in terms of{ti }i=1,...,d as long as the coefficient of si is nonzero for i = 1, . . . , d. ��


3.2 Separable Extensions

In this section we develop criteria for polynomials and extensions to be separa-ble. Recall that an irreducible polynomial f (X) ∈ F[X ] is separable if f (X)

splits into a product of distinct linear factors over a splitting field E. Recallthat a root α of a polynomial f (X) is called simple if X − α divides f (X) but(X − α)2 does not, (otherwise α is called multiple) so an irreducible polyno-mial f (X) is separable if and only if all of its roots in a splitting field E aresimple.

For a polynomial f (X) = ∑ni=0 ci Xi , we let f ′(X) be its derivative

f ′(X) = ∑ni=1 ici X i−1.

Lemma 3.2.1. Let E be an extension of F, let α be an algebraic element of E,and let f (X) ∈ F[X ] a nonzero polynomial with f (α) = 0. Then α is a simpleroot of f (X) if and only if f ′(α) �= 0.

Proof. Write f (X) = (X − α)g(X), a factorization in E[X ]. Then, by theusual rules for derivatives, f ′(X) = (X − α)g′(X) + g(X) so f ′(α) = g(α).

By the division algorithm, we may write g(X) = (X − α)h(X) + g(α),so here g(X) = (X − α)h(X) + f ′(α) and then f (X) = (X − α)g(X) =(X − α)2h(X) + (X − α) f ′(α). Thus (X − α)2 divides f (X) if and only iff ′(α) = 0. ��Proposition 3.2.2. Let f (X) ∈ F[X ] be irreducible. If f ′(X) �= 0, then f (X)

is separable. In particular:(1) If char(F) = 0, then f (X) is separable.(2) If char(F) = p, then f (X) is separable or f (X) is of the form

f (X) =k∑

i=0

ci Xip.

Proof. Let α be any root of f (X) in an extension field E of F. Since f (X)

is irreducible, f (X) is the unique monic polynomial in F[X ] of lowest degreewith f (α) = 0 (and indeed, f (X) = mα(X)). Thus, if f ′(X) �= 0, then f ′(X)

is a polynomial of lower degree, so f ′(α) �= 0 and α is a simple root of f (X)

by Lemma 3.2.1. Hence every root of f (X) in a splitting field is simple andf (X) is separable.

Now if f (X) is a nonconstant polynomial and char F = 0, then f ′(X) �= 0.If char F = p, then f ′(X) �= 0 unless the exponent of every power of Xappearing in f (X) is a multiple of p, proving the proposition. ��

Recall that an algebraic extension E of F is separable if mα(X) ∈ E[X ] isa separable polynomial for every α ∈ E. Thus we see:

3.2 Separable Extensions 55

Corollary 3.2.3. If char F = 0, every algebraic extension of F is separable. ��Example 3.2.4. Here is an example of an inseparable polynomial and, cor-respondingly, an inseparable extension. Let F = Fp(t), the field of rationalfunctions in the variable t with coefficients in Fp. Let E be the field obtainedby adjoining a root of the polynomial X p − t , E = F(s) with s p = t . Then, inE[X ], X p − t = X p − s p = (X − s)p, so s is not a simple root of X p − t . Notethat E is in fact a splitting field for this polynomial. To complete this examplewe need to show that X p − t is an irreducible polynomial in F[X ], and thisfollows immediately from Lemma 3.2.8 below. �Definition 3.2.5. A field F is a perfect field if every algebraic extension of Fis separable. �

We have a simple criterion for a field to be perfect. In order to state it, recallthat for a field F of characteristic p, we have the Frobenius map : F → Fdefined by (a) = a p. We let Fp = Im() and note that, since is anendomorphism, Fp is a subfield of F. (This is easy to verify directly: 1 = 1p,if c = a p and d = bp, then c + d = (a + b)p and cd = (ab)p, and ifc �= 0, c−1 = (a−1)p.)

Theorem 3.2.6. F is a perfect field if and only if:(1) char(F) = 0, or(2) char(F) = p and F = Fp.

Proof. By Corollary 3.2.3, if char(F) = 0, then F is perfect. Let char(F) = p.First assume F = Fp. Let α ∈ E be inseparable and consider mα(X). By

Proposition 3.2.2 (2), mα(X) must be of the form∑k

i=0 ci Xip.Since F = Fp, for each ci there is a di ∈ F with d p

i = ci . Then

mα(X) =k∑

i=0

ci Xip =k∑

i=0

d pi Xip =

( k∑i=0

di Xi)p

contradicting the irreducibility of mα(X). Hence α is separable and F is per-fect.

Next assume F is perfect. (The argument here parallels Example 3.2.4.)Let a ∈ F, a /∈ Fp and consider f (X) = X p − a ∈ F[X ]. By Lemma 3.2.8below, f (X) is irreducible. Let E = F(α) where α is a root of f (X). Then, inE, α p = a so X p − a = X p − α p = (X − α)p and so α ∈ E is an inseparableelement, contradicting the fact that F is perfect. Hence F = Fp. ��Corollary 3.2.7. Every finite field is perfect.


Proof. Immediate from Corollary 2.1.11 and Theorem 3.2.6. ��Lemma 3.2.8. Let char(F) = p, and let a ∈ F, a /∈ Fp. Then f (X) =X pt − a ∈ F[X ] is irreducible for every t ≥ 1.

Proof. Factor f (X) into monic irreducible polynomials in F[X ], f (X) =g1(X) · · · gk(X). Let E = F(α) with g1(α) = 0. Then g1(X) = mα(X) ∈F[X ] as g1(X) is irreducible.

Now 0 = g1(α) implies 0 = f (α) = α pt − a so a = α ptand

f (X) = X pt − a = X pt − α pt = (X − α)pt ∈ E[X ].For any i , gi (X) divides f (X) in F[X ], and hence gi (X) divides f (X) inE[X ] by Lemma 2.2.7, i.e., gi (X) divides (X − α)pt

in E[X ], so gi (X) is apower of X − α and hence gi (α) = 0. Since gi (α) = 0, gi (X) is divisible bymα(X) = g1(X), for every i . Since the choice of g1(X) was arbitrary, we seethat g1(X) = g2(X) = · · · = gk(X) and further that f (X) = g(X)k whereg(X) = (X − α) j for some j .

Thus X pt − a = ((X − α) j )k ∈ E[X ] so j = ps and k = pt−s for somes ≤ t . Suppose s ≤ t − 1. Then g(X) = (X − α)ps ∈ F[X ], so g(X)pt−s−1 ∈F[X ] as well. But g(X)pt−s−1 = (X − α)pt−1 = X pt−1 − b with b = α pt−1

andhence b ∈ F. But then bp = (α pt−1

)p = α p = a, so a ∈ Fp, a contradiction.Hence s = t , k = 1, and f (X) = g(X) is irreducible in F[X ]. ��

3.3 Finite Fields

Let us use the Fundamental Theorem of Galois Theory to completely deter-mine the structure of finite fields. Recall we defined the Frobenius map inDefinition 2.1.10: Let F be a field of characteristic p. Then : F → F is themap defined by (a) = a p.

Theorem 3.3.1. Let p be a prime and r a positive integer.(1) There is a field Fpr of pr elements, unique up to isomorphism.(2) Let s divide r . Then Fpr contains a unique subfield of ps elements. If s

does not divide r , then Fpr does not contain such a field.(3) If E = Fpr and F = Fps for s dividing r, then Gal(E/F) is cyclic of

order r/s, generated by � = s , the Frobenius map. (Then �(α) = α ps.)

In particular, if E = Fpr and F = Fp, then Gal(E/F) is cyclic of order r ,generated by the Frobenius map .

3.3 Finite Fields 57

Proof. (1) Set q = pr . Let f (X) = Xq − X ∈ Fp[X ], and let F be the splittingfield of this polynomial. We know that a splitting field exists and is unique upto isomorphism (Lemma 2.5.2 and Corollary 2.6.4). Let g(X) = Xq−1 − 1, sof (X) = Xg(X). Now f ′(X) = −1 so, by Lemma 3.2.1, the roots of f (X) areall simple. Thus f (X) has deg f (X) = q distinct roots in F, so in particularF must have at least q elements. Now let D ⊆ F be the subset of F consistingof the roots of f (X) and note that |D| = deg f (X) = q. Now we have thatD = {a ∈ F | aq = a} and we observe that D is a field: 1q = 1, if aq = a andbq = b, then (ab)q = ab and (a + b)q = aq + bq = a + b, and for a �= 0,(a−1)q = a−1. Thus f (X) splits in D and since F is a splitting field, F ⊆ D,and so F = D. Hence F = D = Fq is a field of q elements.

On the other hand, if B is any field of q elements, then the multiplicativegroup of B has order q − 1, so every nonzero element of B is a root of g(X) =Xq−1 − 1; hence every element of B is a root of f (X), so B is a splitting fieldof f (X) (again as f (X) cannot split in any proper subfield of D) and hence Bis isomorphic to F.

We have observed that the roots of f (X) are all simple, and hence f (X) isseparable. (Alternatively, this follows from from Proposition 3.2.2 (2).) NowFq is the splitting field of the separable polynomial f (X) ∈ Fp[X ], so Fq isa Galois extension of Fp (Theorem 2.7.14) and | Gal(Fq/Fp)| = (Fq/Fp) =dimFp Fq = r . Now : Fq → Fq and | Fp = id as a p = a for every a ∈ Fp

(by Fermat’s Little Theorem, see Corollary 2.1.12). Hence ∈ Gal(Fq/Fp).

Also, r (α) = α pr = α for every α ∈ F, as every α ∈ Fq is a root off (X), i.e., r = id : Fq → Fq . On the other hand, s �= id for s < r ass(α) = α ps = α if and only if α is a root of the polynomial X ps − X , and thispolynomial cannot have pr roots by Lemma 2.2.1. Hence {1, , . . . , r−1} is asubgroup of Gal(Fq/Fp). Since this subgroup has r = | Gal(Fq/Fp)| elements,it must be the entire Galois group.

Now we may prove (2) and (3) by a direct application of the FundamentalTheorem of Galois Theory. The subfields of Fq are in 1 − 1 correspondencewith the subgroups of the cyclic group {1, , . . . , r−1} and each subgroup isa cyclic group, of order r/s, generated by � = s for some divisor s of r .Thus these subfields are the fixed fields of the cyclic subgroups generated by�, i.e., the fixed fields of �. But �(α) = α is the equation α ps = α, whosesolutions are the roots of X ps − X , and these are the fields Fps as in (1). ��Remark 3.3.2. Theorem 3.3.1 (2) has an alternate, more elementary proof.Consider the multiplicative group G of Fpr , a group of order pr − 1. Sup-pose Fpr contains a subfield isomorphic to Fps , and let H be the multiplicativegroup of that field, of order ps −1. Now, by elementary number theory, ps −1divides pr − 1 if and only if s divides r . Hence if s does not divide r , |H |


does not divide |G|, so no subgroup H , and hence no such subfield Fps canexist. On the other hand, we also know, by Corollary 2.2.2, that G is cyclic,so if s divides r , and hence ps − 1 divides pr − 1, there is a unique subgroupH of order ps − 1, and then H ∪ {0} is the unique subfield of Fpr with ps

elements. �Corollary 3.3.3. Let F = Fps and let f (X) ∈ F[X ] be an irreducible poly-nomial of degree n. Let r = ns. Then E = Fpr is a splitting field for f (X).Furthermore, Gal(E/F) = Z/nZ generated by � = s .

Proof. Let E0 = F(α) be the field obtained by adjoining a single root α off (X) to F. Then, since deg f (X) = n, (E0/F) = n, and E0 = Fpr .

Let E ⊇ E0 be a splitting field for f (X). We claim that E = E0.On the one hand, this claim follows from Theorem 3.3.1. By Theorem

3.3.1, E0 is a Galois extension of F, so by Theorem 2.7.14, it is in particular anormal extension of F. But f (X) has a root in E0; by the definition of a normalextension this implies that f (X) splits in E0.

On the other hand, we can verify this claim directly here. � is an automor-phism of E0 = F(α) of order n, and that implies that α, �(α), . . . , �n−1(α)

are all distinct. But for any k, 0 = �k(0) = �k( f (α)) = f (�k(α)), sof (X) has the n distinct roots α, �(α), . . . , �n−1(α) in E0; therefore f (X) =∏n−1

k=0(X − �k(α)) ∈ E0[X ] is a product of linear factors, i.e., f (X) splits inE0.

The last sentence in the statement of the corollary is then simply a restate-ment of Theorem 3.3.1 (3). ��Corollary 3.3.4. (1) For any p, s, and n, there is an irreducible monic poly-nomial f (X) ∈ Fps [X ] of degree n.

(2) Let k(ps, n) be the cardinality of the set

K = {1 ≤ k ≤ pns − 1 | (pns − 1)/ gcd(k, pns − 1) does not divide

pms − 1 for any proper divisor m of n}.Then the number of distinct irreducible monic polynomials in Fps [X ] of

degree n is k(ps, n)/n.

Proof. (1) By Corollary 2.2.2, the multiplicative group G of Fpns is cyclic. Letα0 be a generator of G, so Fpns = Fps (α0). Then f (X) = mα0(X) ∈ Fps [X ] isan irreducible monic polynomial of degree n.

(2) First we claim that k(ps, n) is the cardinality of the set

K′ = {α ∈ Fpns | α /∈ any proper subfield of Fpns containing Fps }.

3.3 Finite Fields 59

To see this, consider a nonzero element α ∈ Fpns . Then, as an element ofG, α has order dividing pns − 1. By Theorem 3.3.1, the proper subfields ofFpns containing Fps are the fields Fpms with m a proper divisor of n, and α isin such a subfield if and only if its order, as an element of G, divides pms − 1.Now α = αk

0 for some k, and then α has order (pns −1)/ gcd(k, pns −1). Thuswe see that K′ = {αk

0 | k ∈ K}.Now observe that Fps (α) = Fpns if and only if α ∈ K′, and also that

Fps (α) = Fpns if and only if deg mα(X) = n. In this case f (X) = mα(X)

is an irreducible monic polynomial of degree n, and every irreducible monicpolynomial of degree n arises in this way. Also, every such polynomial hasdistinct roots (as it divides X pns − X , which has distinct roots), and no two dis-tinct such polynomials have a root in common (since if f1(α) = f2(α) = 0,then f1(X) = mα(X) = f2(X)). Since each such polynomial has n distinctroots in K′, and K′ has k(ps, n) elements, there are k(ps, n)/n such polyno-mials. ��

Now let us investigate k(ps, n) a bit further.

Lemma 3.3.5. (1) k(ps, n) ≥ ϕ(pns − 1), where ϕ denotes the Euler totientfunction.

(2) If n is prime, k(ps, n) = pns − ps.(3) For any positive integer n,

k(ps, n) =∑d|n

μ(n/d)psd ,

where μ is the Mobius μ function, defined by μ( j) = 0 if j is divisible bythe square of a prime, and otherwise μ( j) = (−1)i where i is the number ofdistinct prime factors of j .

Proof. (1) Observe that k ∈ K for any 1 ≤ k ≤ pns −1 with gcd(k, pns −1) =1, and by definition ϕ(pns − 1) is the number of such integers.

(2) If n is a prime, the only fields intermediate between Fps and Fpns arethese two fields themselves, so in this case K′ = {α ∈ Fpns , α /∈ Fps } andthere are pns − ps such elements.

(3) Let f (ps, n) = pns denote the cardinality of the field Fpns . Then

f (ps, n) =∑d|n

k(ps, d)

as any element α of Fpns generates Fpds for exactly one value of d dividing n.(This is the value of d for which α ∈ Fpds but α /∈ Fpcs for any proper divisor c


of d.) But then the Mobius Inversion Formula from elementary number theoryshows that

k(ps, n) =∑d|n

μ(n/d) f (ps, d),

as claimed. ��

3.4 Disjoint Extensions

In this section, we assume that all fields are subfields of some field A.

Definition 3.4.1. Let B1 and B2 be extensions of F. Then B1 and B2 are dis-joint extensions of F if B1 ∩ B2 = F. �

Of course, B1 and B2 are always disjoint extensions of B1 ∩ B2.We first note the following case, which is very special, but easy and useful.

Note also that in this case we can strengthen Lemma 2.3.4.

Lemma 3.4.2. Let B1 and B2 be finite extensions of F, with di = (Bi/F), i =1, 2. If d1 and d2 are relatively prime, then B1 and B2 are disjoint extensionsof F, and, furthermore,

(B1B2/F) = d1d2, (B1B2/B2) = d1, and (B1B2/B1) = d2.

Proof. Let B0 = B1 ∩ B2 and d0 = (B0/F). Then (Bi/F) = (Bi/B0)(B0/F),for i = 1, 2, so d0 divides both d1 and d2, so d0 = 1 and B0 = F.

Now let d = (B1B2/F). We observe that d = (B1B2/Bi )(Bi/F) fori = 1, 2. From this we immediately see that d is divisible by both d1 andd2, and from Lemma 2.3.4 we see that d ≤ d1d2, so d = d1d2 from which theremaining two equalities immediately follow. ��

As we shall now see, we can say quite a bit about the composite of disjointextensions when at least one of the extensions is finite Galois, and even morewhen both are.

Theorem 3.4.3. Suppose that B2 is a finite Galois extension of B1 ∩ B2. ThenB1B2 is a finite Galois extension of B1, and

Gal(B1B2/B1) ∼= Gal(B2/B1 ∩ B2),

with the isomorphism given by σ �→ σ | B2.

3.4 Disjoint Extensions 61

Proof. Let B0 = B1∩B2. Since B2 is a Galois extension of B0, it is the splittingfield of a separable polynomial f (X) ∈ B0[X ]. If f (X) has roots α1, . . . , αk ,then B2 = B0(α1, . . . , αk) (Remark 2.5.6), and so B1B2 = B1(α1, . . . , αk).Then B1B2 is the splitting field of the separable polynomial f (X) ∈ B1[X ], soB1B2 is a finite Galois extension of B1.

Let σ ∈ Gal(B1B2/B1). Then σ | B1 = id, so certainly σ | B0 = id,and hence σ( f (X)) = f (X). In particular, σ permutes the roots α1, . . . , αk

of f (X), so σ(B2) = B0(σ (α1) . . . , σ (αk)) = B0(α′1, . . . , α

′k) = B2, where

{α′1, . . . , α

′k} is a permutation of {α1, . . . , αk}. Thus σ | B2 : B2 → B2, and

σ | B0 = id, so σ | B2 ∈ Gal(B2/B0).We claim that the map σ �→ σ | B2 is an isomorphism.First, suppose that σ | B2 = id. Then σ(αi ) = αi for each i . Also, since

σ ∈ Gal(B1B2/B1), σ | B1 = id by definition. But we have observed thatB1B2 = B1(α1, . . . , αk), so σ = id.

Next, let τ ∈ Gal(B2/B0). Then B1B2 = B1(α1, . . . , αk) is the splittingfield of f (X) ∈ B1[X ], so, by Lemma 2.6.3, τ extends to σ : B1B2 → B1B2,and then σ | B2 = τ . ��

Under our hypothesis we may strengthen Lemma 2.3.4.

Corollary 3.4.4. Suppose that B2 is a finite Galois extension of B1 ∩ B2. Then

(B1B2/B1) = (B2/B1 ∩ B2),

(B1B2/B2) = (B1/B1 ∩ B2),

and

(B1B2/B1 ∩ B2) = (B1/B1 ∩ B2)(B2/B1 ∩ B2).

Proof. The first equality is immediate from Theorem 3.4.3 and then the secondand third follow as

(B1B2/B1 ∩ B2) = (B1B2/B1)(B1/B1 ∩ B2)

= (B1B2/B2)(B2/B1 ∩ B2). ��We also make the following observations.

Corollary 3.4.5. Let B1 and B2 be finite extensions of F and suppose that B2

is a Galois extension of F. Then

(B1B2/F) = (B1/F)(B2/F)

if and only if B1 and B2 are disjoint extensions of F.


Proof. By Corollary 3.4.4,

(B1/F)(B2/F) = (B1/B1 ∩ B2)(B1 ∩ B2/F)(B2/B1 ∩ B2)(B1 ∩ B2/F)

= (B1B2/B1 ∩ B2)(B1 ∩ B2/F)2

= (B1B2/F)(B1 ∩ B2/F)

so we have the equality in the statement of the corollary if and only if(B1 ∩ B2/F) = 1, i.e., if and only if B1 ∩ B2 = F. ��Corollary 3.4.6 (Theorem on Natural Irrationalities). Let f (X) ∈ F[X ] bea separable polynomial and let B ⊇ F. Let E be a splitting field for f (X) overF. Then BE is a splitting field for f (X) over B and Gal(BE/B) is isomorphicto Gal(E/E ∩ B), a subgroup of Gal(E/F), with the isomorphism given byσ �→ σ | E.

Proof. Let f (X) have roots α1, . . . , αk , so E = F(α1, . . . , αk) (Remark 2.5.6);then BE = B(α1, . . . , αk) is a splitting field for f (X) ∈ B[X ]. Now BE is aGalois extension of B as it is the splitting field of a separable polynomial, andE is a Galois extension of F for the same reason. Then, setting D = B ∩ E,F ⊆ D ⊆ E, E = D(α1, . . . , αk), and so E is a Galois extension of D, also forthe same reason. Then, by Theorem 3.4.3, Gal(BE/B) is isomorphic, via therestriction map, to Gal(E/D), which is a subgroup of Gal(E/F). ��Theorem 3.4.7. (1) Let B1 and B2 be disjoint Galois extensions of F. ThenE = B1B2 is a Galois extension of F with Gal(E/F) = Gal(B1/F) ×Gal(B2/F).

(2) Let E be a Galois extension of F and suppose that Gal(E/F) = G1 ×G2. Let B1 = Fix(G1) and B2 = Fix(G2). Then B1 and B2 are disjoint Galoisextensions of F and E = B1B2.

Proof. (1) By Theorem 3.4.3, we have isomorphisms

Gal(E/Bi ) → Gal(B j/F)

given by σ �→ σ | B j , for i = 1 and j = 2, and vice versa. In other words, anyσ j ∈ Gal(B j/F) has a unique extension to σ j ∈ Gal(E/Bi ) ⊆ Gal(E/F), j =1, 2. Now any e ∈ E may be written as e = ∑

b1i b2

i with b1i ∈ B1 and b2

i ∈ B2.Then σ1σ2(e) = ∑

σ1σ2(b1i b2

i ) = ∑σ1σ2(b1

i )σ1σ2(b2i ) = ∑

σ1(b1i )σ2(b2

i ) =∑σ2σ1(b1

i )σ2σ1(b2i ) = ∑

σ2σ1(b1i b2

i ) = σ2σ1(e), so σ1 and σ2 commute andwe have a map

Gal(B1/F) × Gal(B2/F) �→ Gal(E/F)


defined by

(σ1, σ2) �→ σ1σ2.

Next, this map is an injection: Suppose σ1σ2 = id : E → E. Then inparticular σ1σ2 | B2 = id, but σ2 | B2 = id by definition, so σ1 = σ1 | B1 = id,and similarly σ2 = σ2 | B2 = id.

Finally, on the one hand, as Gal(E/F) contains Gal(B1/F) × Gal(B2/F),

| Gal(E/F)| ≥ | Gal(B1/F) × Gal(B2/F)|= | Gal(B1/F)| | Gal(B2/F) |= (B1/F)(B2/F)

= (B1/F)(E/B1)

= (E/F),

and, on the other hand, | Gal(E/F)| = (E/D) ≤ (E/F), where D ⊇ F is thefixed field of Gal(E/F), so | Gal(E/F)| = (E/F). Thus this map is an injectionfrom one group to another group of the same order, and hence an isomorphism;furthermore E is a Galois extension of F.

(2) Since G = G1G2 and {1} = G1 ∩ G2, this follows directly fromProposition 2.8.14. ��Corollary 3.4.8. Let B1 and B2 be finite Galois extensions of F and let E =B1B2, B0 = B1 ∩ B2. Then E is a Galois extension of F and Gal(E/F) ={(σ1, σ2) ∈ Gal(B1/F) × Gal(B2/F) | σ1 | B0 = σ2 | B0}.Proof. Since B1 is a Galois extension of F, it is the splitting field of a sep-arable polynomial f1(X) ∈ F[X ], by Theorem 2.7.14. Similarly, B2 is thesplitting field of a separable polynomial f2(X) ∈ F[X ]. Then E is the split-ting field of the separable polynomial f (X) = f1(X) f2(X) ∈ F[X ], so, byTheorem 2.7.14, it is a Galois extension of F.

As in the proof of Theorem 3.4.3, σ(Bi ) = Bi for each σ ∈ Gal(E/F),i = 1, 2 (as Bi is the splitting field of a separable polynomial whose roots arepermuted by σ ). Thus we have a map σ �→ (σ1, σ2) = (σ | B1, σ | B2).Again this map is an injection, as E = B1B2, and clearly any (σ1, σ2) in imageof this map satisfies σ1 | B0 = σ2 | B0. Then

(E/F) = (E/B0)(B0/F) = (E/B2)(B2/B0)(B0/F)

= (B1/B0)(B2/B0)(B0/F)

= (B1/F)(B2/F)/(B0/F),

and these are the orders of the groups in the statement of the corollary. ��


Corollary 3.4.9. Let B and E be disjoint extensions of F, with E a finite Galoisextension of F. Let r : Gal(BE/B) → Gal(E/F) be the restriction isomor-phism, r(σ ) = σ | E. Let H ⊆ Gal(BE/B) be a subgroup. Then, for the fixedfields, we have

Fix(H) = B Fix(r(H)).

Proof. First observe that r is an isomorphism by Theorem 3.4.3. Now let K =r(H) and D = Fix(K ). Then certainly Fix(H) ⊇ BD. Now

(BE/B) = (BE/BD)(BD/B)

and

(E/F) = (E/D)(D/F).

Now (BE/B) = (E/F) as the Galois groups of these two extensions areisomorphic, again by Theorem 3.4.3. On the other hand, by Lemma 2.3.4,

(BE/BD) ≤ (E/D) and (BD/B) ≤ (D/F),

so both inequalities must be equalities. But, since r is an isomorphism betweenthe groups H and K ,

(BE/ Fix(H)) = |H | = |K | = (E/ Fix(K )) = (E/D)

so (BE/ Fix(H)) = (BE/BD) and hence Fix(H) = BD. ��Theorem 3.4.10. Let B and E be disjoint extensions of F.

(1) Assume that both B and E are finite Galois extensions of F. Let N1 =Gal(BE/B) and N2 = Gal(BE/E). Then BE is a Galois extension of F, andN1 and N2 are normal subgroups of Gal(BE/F). Furthermore, Gal(BE/F) isthe direct product N1 × N2 of the subgroups N1 and N2.

(2) Assume that E is a finite Galois extension of F. Let H = Gal(BE/B)

and N = Gal(BE/E). Then BE is a Galois extension of F and N is a normalsubgroup of Gal(BE/F). Furthermore, Gal(BE/F) is the semidirect productH N of the subgroups H and N.

Proof. (1) This follows immediately from Theorem 3.4.3 and Corollary 3.4.6.(2) E is a Galois extension of F, by hypothesis, so N = Gal(BE/E) is a

normal subgroup of Gal(BE/F), with quotient H0 = Gal(E/F). By Theorem3.4.3 the subgroup H = Gal(BE/B) maps isomorphically onto H0 under theprojection G → G/N . But then G = H N . ��


Lemma 3.4.11. Let α ∈ A with separable minimum polynomial mα(X) ∈F[X ]. Let E be a splitting field of mα(X). If B is an extension of F such that Band E are disjoint, then mα(X) is irreducible in B[X ].Proof. Let mα(X) be the minimum polynomial of α over B. Of course, mα(X)

is irreducible in B[X ] (as is mα(X) in F[X ]). Let {α1, . . . , αd} be the roots ofmα(X) in E. E is a Galois extension of F by Theorem 2.7.14. Now Gal(E/F)

acts transitively on {αi }, by Proposition 2.8.15, and every σ0 ∈ Gal(E/F) ex-tends to σ ∈ Gal(EB/B), by Theorem 3.4.3, so Gal(EB/B) acts transitivelyon {αi }. Then, by Lemma 2.7.12, mα(X) = ∏d

i=1(X − αi ) = mα(X). ��Remark 3.4.12. The converse of Lemma 3.4.11 is false in general. Here is anexample. Let F = Q and A = Q(ω,

3√

2) (ω a primitive cube root of 1).Let α = 3

√2, so mα(X) = X3 − 2 ∈ F[X ] with splitting field E = A. Let

B = Q(ω). Then mα(X) is irreducible in B[X ], but B ∩ E = B ⊃ F. �However, we do have the following partial converse.

Lemma 3.4.13. Let α ∈ A with separable minimum polynomial mα(X) ∈F[X ] and suppose that E = F(α) is a splitting field of mα(X). If B is an ex-tension of F such that mα(X) is irreducible in B[X ], then B and E are disjointextensions of F.

Proof. Since mα(X) is irreducible in B[X ], mα(X) = mα(X). By Corollary3.4.4, (B(α)/B) = (BF(α)/B) = (F(α)/B ∩ F(α)) = (E/B ∩ E). But

(B(α)/B) = deg mα(X) = deg mα(X) = (F(α)/F) = (E/F),

so B ∩ E = F. ��Let B1 and B2 be extensions of F. In Lemma 2.3.4 we showed that

(B1B2/F) ≤ (B1/F)(B2/F). In Example 2.3.7 we gave an example withB1 and B2 disjoint extensions of F where we had strict inequality. We nowformulate a condition that ensures that we will have equality. Note that inLemma 3.4.14 and Definition 3.4.15 we do not restrict ourselves to finite ex-tensions.

Lemma 3.4.14. Let B1 and B2 be extensions of F. The following are equiva-lent:

(a) If S1 = {s ′i } is a subset of B1 that is F-linearly independent, then S1 is

B2-linearly independent.(b) If S2 = {s ′′

j } is a subset of B2 that is F-linearly independent, then S2 isB1-linearly independent.

(c) If S1 = {s ′i } is a subset of B1 that is F-linearly independent, and S2 =

{s ′′j } is a subset of B2 that is F-linearly independent, then S1S2 = {s ′

i s′′j } is

F-linearly independent.


Proof. We show that (a) and (c) are equivalent. Then (b) and (c) are equivalentsimply by interchanging the roles of B1 and B2, so all three are equivalent.

(a) implies (c): Suppose∑

i, j ai j s ′i s

′′j = 0 with ai j ∈ F. Then(∑

i s ′i (

∑j ai j s ′′

j )) = 0. As S1 is B2-linearly independent, each inner sum is

zero; as S2 is F-linearly independent, each ai j is zero.(c) implies (a): Suppose

∑i βi s ′

i = 0 with βi ∈ B2. Choose S2 = {s ′′j } to

be a basis of B2 as an F-vector space. Then, for each i , we have βi = ∑j ai j s ′′

jwith ai j ∈ F. Then 0 = ∑

i βi s ′i = ∑

i s ′i (

∑j ai j s ′′

j ) = ∑i, j ai j s ′

i s′′j . As

S1S2 = {s ′i s

′′j } is F-linearly independent, each ai j is zero, and hence each βi is

zero. ��Definition 3.4.15. Let B1 and B2 be extensions of F. If B1 and B2 satisfy theequivalent conditions of Lemma 3.4.14, then B1 and B2 are linearly disjointextensions of F. �Lemma 3.4.16. Let B1 and B2 be finite extensions of F. Then (B1B2/F) =(B1/F)(B2/F) if and only if B1 and B2 are linearly disjoint extensions of F.

Proof. Let S1 be a basis for B1 and S2 be a basis for B1. Then S1S2 spansB1B2 as an F-vector space. If B1 and B2 are linearly disjoint then S1S2 islinearly independent and hence a basis for B1B2 over F. Thus in this case(B1B2/F) = (B1/F)(B2/F). If B1 and B2 are not linearly disjoint then S1S2 isnot linearly independent and hence some proper subset of S1S2 is a basis forB1B2 over F. Thus in this case (B1B2/F) < (B1/F)(B2/F). ��Corollary 3.4.17. Let B1 and B2 be disjoint finite extensions of F and sup-pose that B2 is a Galois extension of F. Then B1 and B2 are linearly disjointextensions of F.

Proof. Immediate from Corollary 3.4.5 and Lemma 3.4.16. ��

3.5 Simple Extensions

Let E be a finite extension of F. Then, choosing a basis {α1, . . . , αk} for E asan F-vector space, E = F(α1, . . . , αk). We may think of E as obtained from Fin (at most) k steps, successively adjoining α1, then α2, . . . , and finally αk . Itis natural to ask when we can obtain E in one step.

Definition 3.5.1. Let E be an algebraic extension of F such that E = F(α) forsome α in E. Then E is a simple extension of F and α is a primitive elementof E. �

3.5 Simple Extensions 67

We derive a criterion for E to be a simple extension of F.

Theorem 3.5.2. Let E be a finite extension of F. Then E is a simple extensionof F if and only if there are a finite number of intermediate fields F ⊆ B ⊆ E.

Proof. First, we consider the case when F is finite. Then E is finite, so on theone hand there are (trivially) only finitely many intermediate fields, and onthe other hand, the multiplicative group of E is cyclic (Corollary 2.2.2), so itconsists of the powers of some α ∈ E, and then E = F(α).

Now we consider the case when F is infinite. First, assume that E is simple,E = F(α). Then α has minimum polynomial mα(X) ∈ F[X ]. Let F ⊆ B ⊆ Eand consider mα(X) ∈ B[X ], the minimum polynomial of α over B. Then E =B(α), so (E/B) = deg mα(X). Let D be the field obtained from F by adjoiningthe coefficients of mα(X). Then F ⊆ D ⊆ B. As mα(X) is irreducible in B[X ],it is certainly irreducible in D[X ], so mα(X) is the minimum polynomial of α

over D, and E = D(α), so (E/D) = deg mα(X) = (E/B) and hence B = D.Thus the intermediate field B is determined by the polynomial mα(X). Butmα(X) divides mα(X) in B[X ] (as mα(α) = 0) and hence mα(X) dividesmα(X) in E[X ]. But mα(X) has only finitely many factors in E[X ], so thereare only finitely many possibilities for mα(X) and hence for B.

Conversely, suppose there are only finitely many fields between F and E.We show that given any α and β in E there is a γ1 in E with F(α, β) = F(γ1).Since E is a finite extension of F, it is obtained by adjoining a finite number kof elements, and then the theorem follows by induction on k.

Consider the fields F(α + aβ) for a ∈ F. Since there are infinitely manyelements of F and only finitely many intermediate fields, there must exist dis-tinct elements a1 and a2 of F with F(α+a1β) = F(α+a2β). Set γ1 = α+a1β,γ2 = α+a2β. Then F(γ1) = F(γ2), so γ2 ∈ F(γ1), and certainly γ1 ∈ F(γ1), soγ2−γ1 = (a2−a1)β ∈ F(γ1), and hence β ∈ F(γ1) and γ1−a1β = α ∈ F(γ1).Thus F(γ1) ⊆ F(α, β) ⊆ F(γ1) and they are equal. ��

While Theorem 3.5.2 does not look easy to apply, in fact it readily gives astrong general result.

Corollary 3.5.3. If E is a finite separable extension of F, then E is a simpleextension of F.

Proof. Let E = F(α1, . . . , αk). By the definition of a separable extension(Definition 2.7.3) each mαi (X) is a separable polynomial and hence (also byDefinition 2.7.3) the product f (X) = mα1(X) · · · mαk (X) is also a separablepolynomial. Let E′ ⊇ E be an extension of F that is a splitting field of thepolynomial f (X). Then E′ is a Galois extension of F (Theorem 2.7.14), so by


the Fundamental Theorem of Galois Theory there are only finitely many fieldsintermediate between F and E′ (corresponding to the subgroups of Gal(E′/F)),and hence there are certainly only finitely many fields between F and E. Then,by Theorem 3.5.2, E is a simple extension of F. ��

In case E is a Galois extension of F, we have the following criterion for anelement of E to be primitive.

Lemma 3.5.4. Let E be a finite Galois extension of F and let α ∈ E. Then α

is primitive, i.e., E = F(α), if and only if σ(α) �= α for every σ ∈ Gal(E/F),σ �= id.

First proof. Let d = (F(α)/F) and n = (E/F), so E = F(α) if and onlyif d = n. Now n = | Gal(E/F)| (Theorem 2.8.5) and d = deg mα(X)

(Lemma 2.4.6 (2)). But, by Lemma 2.7.12, mα(X) = ∏di=1(X − αi ) where

α1 = α, α2, . . . , αd are the Galois conjugates of α in E. Hence E = F(α)

if and only if α has n conjugates, and this is true if and only if σ(α) = α,σ ∈ Gal(E/F), implies σ = id.

Second Proof. Let B = F(α) so B is intermediate between F and E. By theFTGT, F = Fix(H) for H a subgroup of Gal(E/F), and furthermore H ={σ ∈ Gal(E/F) | σ | B = id} = {σ ∈ Gal(E/F) | σ(α) = α}. Then, also bythe FTGT, B = E if and only if H = {id}. ��

The following proposition is very useful in constructing primitive ele-ments.

Proposition 3.5.5. Let F be a field of characteristic 0 and let B1 and B2 bedisjoint finite Galois extensions of F. Set B = B1B2. Let α1 be a primitiveelement of B1 and let α2 be a primitive element of B2. Then α = α1 + α2 is aprimitive element of B.

Proof. We first observe that B is a Galois extension of F (Theorem 3.4.7), so,by Lemma 3.5.4, it suffices to show that the Galois conjugates of α are alldistinct. In order to show this, we need only show that if σ(α) = α for someσ ∈ G = Gal(B/F), then σ = id.

Thus, let σ ∈ G with σ(α) = α. By Theorem 3.4.7, σ = (σ1, σ2) withσi ∈ Gal(Bi/F), i = 1, 2. Then

α1 + α2 = α = σ(α) = σ1(α1) + σ2(α2),

so we see that σ1(α1) − α1 = σ2(α2) − α2. But σi (αi ) − αi ∈ Bi , i = 1, 2, andB1 and B2 are disjoint extensions of F, so σ1(α1)−α1 = σ2(α2)−α2 = a ∈ F.Thus σ1(α) = α + a, from which we see that σ k

1 (α) = α + ka for every k.

3.6 The Normal Basis Theorem 69

But σ1 has finite order m, so α = σ m1 (α) = α + ma, so ma = 0 and hence

a = 0. Thus σ1(α1) = α1, and then σ2(α2) = α2 as well. But α1 is a primitiveelement of B1, so σ1(α1) = α1 implies σ1 = id on B1; similarly σ2 = id on B2

as well, so σ = id on B as required. ��Remark 3.5.6. In order to effectively apply Proposition 3.5.5 we need to de-velop some more background, which we do in the next chapter. But to showits applicability, as well as to provide some interesting examples of primitiveelements, we shall quote an example from the next chapter here.

Example 4.6.21. (1) Let E = Q(√

2,3√

2,5√

2,7√

2). Then E is an ex-tension of Q of degree 210 with primitive element

√2+ 3

√2+ 5

√2+ 7

√2.

(2) Let E = Q(√

2,√

3,√

5,√

7). Then E is an extension of Q ofdegree 16 with primitive element

√2 + √

3 + √5 + √

7. Similarly, ifE = Q(

3√

2,3√

3,3√

5,3√

7), then E is an extension of Q of degree 81with primitive element 3

√2 + 3

√3 + 3

√5 + 3

√7.

(3) Let E = Q(√

2,3√

2,√

3,3√

3). Then E is an extension of Q ofdegree 36 with primitive element

√2 + 3

√2 + √

3 + 3√

3. �Example 3.5.7. Here is an example that shows that the conclusion of Proposi-tion 3.5.5 may be false if B1 and B2 are not Galois extensions of F. Let F = Q,B1 = Q(

3√

2), and B2 = Q(ω3√

2). Then B = Q(ω,3√

2). Let α1 = − 3√

2, aprimitive element of B1, and let α2 = −ω

3√

2, a primitive element of B2. Thenα = α1 + α2 = ω2 3

√2 is not a primitive element of B. �

Example 3.5.8. Here is an example of an extension that is not simple. LetF = Fp(s, t), the field of rational functions in the variables s and t over Fp.Let E be the splitting field of the polynomial (X p − s)(X p − t). More simply,E = F(β, γ ) where β p = s and γ p = t . Then [E/F] = p2. On the other hand,it is easy to check that for any α ∈ E, α p ∈ F, so [F(α) : F] = 1 or p, and inparticular F(α) ⊂ E. �

3.6 The Normal Basis Theorem

Let E be a Galois extension of F of degree n. Then, as an F-vector space,E has a basis of n elements. On the other hand, Gal(E/F) has n elements,and so it natural to ask if there is some element α of E whose conjugates{σ(α) | σ ∈ Gal(E/F)} form a basis for E over F. In fact, this is always thecase, as we shall show in this section.

Definition 3.6.1. Let E be a Galois extension of F. A basis {αi }i=1,...,n of E asa vector space over F is a normal basis for E over F if {αi } = {σi (α)} for someα ∈ E, where {σi }i=1,...,n = Gal(E/F). �


There is one case of this theorem that follows directly from linear algebra.

Theorem 3.6.2. Let E be a finite Galois extension of F. Suppose Gal(E/F) iscyclic. Then E has a normal basis.

Proof. Let σ be a generator of Gal(E/F) and let T : E → E be given byT (α) = σ(α). Then T is a linear transformation that satisfies T n − I = 0. Weclaim the minimum polynomial mT (X) = Xn − 1. Certainly mT (X) dividesXn − 1. Suppose mT (X) �= Xn − 1. Then mT (X) = ∑d

i=0 ai Xi with d < n.In other words,

a0 id + · · · + ad−1σd−1 = 0

as a map from E to E, and hence as a map from E∗ to E.But {id, σ, . . . , σ d−1} are a set of distinct characters of E∗ in E, so by

Theorem 2.8.4 are linearly independent, so ai = 0 for each i , a contradiction.Now order the elements of Gal(E/F) as {id, σ, . . . , σ d−1}.Then (by [AW, Theorem 3.7.1]) there is an element α of E with Ann(α) =

〈mT (X)〉 (i.e., where the annihilator ideal of α in F[X ] is the ideal generatedby mT (X)). Then, since deg mT (X) = n, {α, T (α), . . . , T n−1(α)} ={α, σ (α), . . . , σ n−1(α)} are linearly independent, hence form a basis, and thisis a normal basis.

(Alternatively phrased, mT (X) = mα(X), by Proposition 2.4.6 (3). But Ehas a basis over F in which the matrix of T is the companion matrix of mT (X).But then if α1 is the first element of the basis, σ j (α1) = T j (α1) = α1+ j forj = 0, . . . , n − 1 ([AW, Corollary IV.4.15]), so {αi } is a normal basis.) ��

The argument for the general case also uses linear algebra, but is moreinvolved.

Lemma 3.6.3. Let E be a Galois extension of F of degree n, and let σi , i =1, . . . , n, be the elements of Gal(E/F). Let α1, . . . , αn be elements of E. Then{α1, . . . , αn} is a basis for E over F if and only if the matrix A = (αi j ), αi j =σi (α j ), is nonsingular.

Proof. Let {α1, . . . , αn} be a basis for E over F, and suppose some linearcombination of the rows of A is zero, say

∑ni=1 ciαi j = 0 for each j , i.e.,∑n

i=1 ciσi (α) = 0, α = α1, . . . , α j . But each σi is an F-linear map, so∑ni=1 ciσi (α) = 0 for every α ∈ E. But the σi are distinct characters of E∗

in E, so by Theorem 2.8.4 are linearly independent; so c1 = · · · = cn = 0.Thus the rows of A are linearly independent and A is nonsingular.

On the other, suppose {α1, . . . , αn} is not a basis. Then it is not linearlyindependent, so there is a nontrivial linear combination

∑nj=1 d jα j = 0. Then

3.6 The Normal Basis Theorem 71

∑nj=1 d jαi j = ∑n

j=1 d jσi (α j ) = σi (∑n

j=1 d jα j ) = σi (0) = 0 for each i , sothe columns of A are linearly dependent, and A is singular. ��Lemma 3.6.4. Let F be an infinite field, let E be an extension of F, and letf (X1, . . . , Xk) ∈ E[X1, . . . , Xk] be a polynomial. If f (a1, . . . , ak) = 0 forevery (a1, . . . , ak) ∈ Fk , then f (X1, . . . , Xk) is the zero polynomial.

Proof. By induction on k. For k = 1, let f (X1) ∈ E[X1] be a nonzero polyno-mial with f (a1) = 0 for every a1 ∈ F. Then f (X1) has more than deg f (X1)

roots, contradicting Lemma 2.2.1. Now assume the lemma is true for k −1 andconsider f (X1, . . . , Xk). Regard this as a polynomial in Xk with coefficientsin X1, . . . , Xk−1, so

f (X1, . . . , Xk) =n∑

i=0

gi (X1, . . . , Xk−1)Xik .

For any fixed elements a1, . . . , ak−1 of F, f (Xk) = f (a1, . . . , ak−1, Xk)

has more than d roots, so is identically zero by the k = 1 case, so each coef-ficient gi (X1, . . . , Xk−1) is zero for every (a1, . . . , ak−1), and hence is identi-cally zero by the k − 1 case. ��

Theorem 2.8.4 gives us linear independence of characters. In fact, whenF is infinite and E is a finite Galois extension of F, we may derive a strongerkind of independence, algebraic independence.

Theorem 3.6.5. Let F be an infinite field and let Gal(E/F) = {σi }, i =1, . . . , n. Then {σi } are algebraically independent, i.e., if f (X1, . . . , Xk) ∈E[X1, . . . , Xk] is a polynomial with f (σ1(α), . . . , σk(α)) = 0 for everyα ∈ E, then f (X1, . . . , Xk) is the zero polynomial.

Proof. Fix a basis {α1, . . . , αn} for E over F. By Lemma 3.6.3, the matrixA = A(α1, . . . , αn) is nonsingular, where A = (αi j ) = (σi (α j )).

In this proof we shall identify an n-tuple (β1, . . . , βn) of elements of Ewith the column vector [β1, . . . , βn]t . (Purely for typographical convenience,we write this column vector [β1, . . . , βn]t as the transpose of the row vector[β1, . . . , βn].)

For an arbitrary n-tuple (c1, . . . , cn) of elements of F, let α = ∑nj=1 c jα j .

Then σi (α) = ∑nj=1 c jσi (α j ) = ∑n

j=1 αi j c j for each i = 1, . . . , n, i.e.,

[σ1(α), . . . , σn(α)]t = A[c1, . . . , cn]t .

Now suppose that f (σ1(α), . . . , σn(α)) = 0 for every α ∈ E. Then, in ournotation,


0 = f ([σ1(α), . . . , σn(α)]t ) = f (A[c1, . . . , cn]t )

for every (c1, . . . , cn) ∈ Fn . Thus the polynomial

g(X1, . . . , Xn) = f (A[X1, . . . , Xn]t )

is zero for every (c1, . . . , cn) ∈ Fn , so, by Lemma 3.6.4, it must be the zeropolynomial in E[X1, . . . , Xn], and g(γ1, . . . , γn) = 0 for every (γ1, . . . , γn) ∈En . We claim also that f (X1, . . . , Xn) is the zero polynomial, and, againby Lemma 3.6.4, that will follow if we show f (b1, . . . , bn) = 0 for ev-ery (b1, . . . , bn) ∈ Fn . Thus, let (b1, . . . , bn) ∈ Fn . Since A is invertible,we may define (γ1, . . . , γn) by [γ1, . . . , γn]t = A−1[b1, . . . , bn]t , and thenf (b1, . . . , bn) = f ([b1, . . . , bn]t ) = f (A[γ1, . . . , γn]t ) = g(γ1, . . . , γn) = 0,as claimed. ��Remark 3.6.6. Theorem 3.6.5 is false for finite fields. Here is a counterexam-ple. Let F = F2 = {0, 1} and E = F4 = {0, 1, α0, α0 + 1} with α2

0 = α0 + 1.Let Gal(E/F) = {σ1, σ2} with σ1 = id and σ2 given by σ2(α0) = α + 1.Let f (X1, X2) be the polynomial f (X1, X2) = (X2 − X1)(X2 − X1 − 1).Then p(σ1(α)σ2(α)) = (σ2(α) − σ1(α))(σ2(α) − σ1(α) − 1) = 0 for everyα ∈ E. �Theorem 3.6.7. Let E be a finite Galois extension of F. Then E has a normalbasis.

Proof. If F is finite, then E is a cyclic extension of F (Theorem 3.3.1) so theresult is a special case of Theorem 3.6.2.

Assume henceforth that F is infinite. Let Gal(E/F) = {σ1, . . . , σn}. LetB = B(X1, . . . , Xn) be the matrix whose entries are indeterminates Xk , withbi j = Xk if σiσ j = σk . Then det(B) is a polynomial in X1, . . . , Xk , det(B) =d(X1, . . . , Xk). Note each Xi appears exactly once in every row and column.Thus, setting X1 = 1 and Xi = 0 for i > 1, we see det(B) = ±1, i.e.,d(1, 0, . . . , 0) = ±1. Thus d is not the zero polynomial. Hence by Theorem3.6.5 there is an α ∈ E with d(σ1(α), . . . , σn(α)) �= 0. Let α j = σ j (α),j = 1, . . . , n. Now let A = B(σ1(α), . . . , σn(α)). If A = (ai j ), then ai j = αk

where σiσ j (α) = σk(α), or, in other words, ai j = σiσ j (α) = σi (α j ). Thus Ais a matrix as in Lemma 3.6.3, and det(A) = d(σ1(α), . . . , σn(α)) �= 0, so Ais nonsingular, and then, by Lemma 3.6.3, {α1, . . . , αn} = {σ1(α), . . . , σn(α)}is a basis for E over F. ��Remark 3.6.8. Let E be a Galois extension of F and let α ∈ E. Then α is aprimitive element if the conjugates of α are all distinct (Lemma 3.5.4), but

3.7 Abelian Extensions and Kummer Fields 73

for α to be part of a normal basis, the conjugates of α must be linearly inde-pendent. This latter is definitely a stronger condition, as the following simpleexample shows. Let F = Q and E = Q(

√2). Let α = √

2. Then the conju-gates of α are α and −α, so the conjugates are distinct (and E = F(α)) butthey are certainly not linearly independent (so {α, −α} is not a normal basisfor E over F). �

3.7 Abelian Extensions and Kummer Fields

Let E be a Galois extension of F. Then E is called an abelian extension ofF if the Galois group Gal(E/F) is abelian. Since abelian groups are easierto understand than arbitrary groups, we may hope that abelian extensions areeasier to understand than arbitrary Galois extensions. As we shall see in thissection, under the appropriate conditions on F, we can indeed describe abelianextensions very concretely.

Recall that for a field F, F0 denotes the prime field: F0 = Q if char(F) = 0and F0 = Fp if char(F) = p.

Definition 3.7.1. An element ζ ∈ F is a primitive nth root of 1 in F if ζ n = 1,but ζ m �= 1 for any 1 ≤ m < n. �Lemma 3.7.2. (1) If F has a primitive nth root of 1, then char(F) = 0 or isprime to n.

(2) The following are equivalent:(a) F has a primitive nth root of 1.(b) F has n distinct nth roots of 1.

Proof. (1) Suppose char(F) = p and p divides n. Let ζ n = 1. Then ζ is a rootof Xn −1. But Xn −1 = (Xn/p)p −1 = (Xn/p −1)p so ζ is a root of Xn/p −1,i.e., ζ n/p = 1 and ζ is not primitive.

(2) If ζ is a primitive nth root of 1, then 1, ζ, . . . , ζ n−1 are distinct andare all nth roots of 1. On the other hand, the group of nth roots of 1 in F is asubgroup of the multiplicative group of F, so is cyclic (Corollary 2.2.2). Thusif F has n distinct nth roots of 1, a generator of this group is a primitive nth rootof 1. ��

Henceforth we assume that char(F) = 0 or that p = char(F) is relativelyprime to n and we let ζn denote an arbitrary but fixed primitive nth root of 1.Observe that ζn = exp(2π i/n) is a primitive nth root of 1 in C.

Lemma 3.7.3. F0(ζn) is a Galois extension of F0, and Gal(F0(ζn)/F0) is iso-morphic to a subgroup of (Z/nZ)∗. In particular this group is abelian.


Proof. F(ζn) is the splitting field of the separable polynomial Xn − 1, so it isa Galois extension of F0.

If σ1 ∈ Gal(F0(ζn)/F0), then σ1(ζn) is a primitive nth root of 1, so σ1(ζn) =ζ k1

n for some k1 with (k1, n) = 1.Similarly, σ2 ∈ Gal(F(ζ0)/F) satisfies σ2(ζn) = ζ k2

n where (k2, n) = 1.Then σ1σ2(ζn) = ζ k1k2

n = σ2σ1(ζn), and σ ∈ Gal(F0(ζn)/F0) is determined byσ(ζn); so we see that we have an injection σi �→ ki from Gal(F0(ζn)/F0) into(Z/nZ)∗. ��Remark 3.7.4. We will see in Corollary 4.2.7 below that if F0 = Q, thenGal(F0(ζn)/F0) = (Z/nZ)∗. This is false in general. In the most extremecase, if p−1 is a multiple of n, then ζn ∈ F0, so in this case Gal(F0(ζn)/F0) ={id}. �

We now introduce some nonstandard but very useful language.

Definition 3.7.5. Let α ∈ F. Then α is n-powerless in F if α is not an m th

power in F for any m dividing n, m > 1. �Proposition 3.7.6. Let F ⊇ F0(ζn). Let E be the splitting field of Xn − a ∈F[X ]. Then Gal(E/F) is isomorphic to a subgroup of Z/nZ. In particular, thisgroup is cyclic. Furthermore, the following are equivalent:

(1) a is n-powerless in F.(2) Gal(E/F) ∼= Z/nZ.(3) Xn − a is irreducible in F[X ].

Proof. Let α be a root of Xn −a in E. Then Xn −a = ∏n−1i=0 (X − ζ i

nα) so E =F(α), and mα(X) divides Xn − a in F[X ] and hence in E[X ] (Lemma 2.2.7)so mα(X) = ∏m−1

k=0 (X − ζ ikn α) for some subset {ik} of {0, . . . , n − 1}.

Since ζ ikn α and α are both roots of the irreducible polynomial mα(X), by

Lemma 2.6.1 there is an element σk ∈ Gal(E/F) with σk(α) = ζ ikn α, and

this equation determines σk uniquely. Furthermore, if σ�(α) = ζ i�n α, then

σkσ�(α) = ζ ik+i�n α, so we see that {ik} form a subgroup of Z/nZ isomorphic

to Gal(E/F).

Now to the second part of the proposition. Since | Gal(E/F)| = (E/F) =(F(α)/F) = deg mα(X),

Gal(E/F) = Z/nZ ⇔ | Gal(E/F)| = n

⇔ deg mα(X) = n

⇔ mα(X) = Xn − a

⇔ Xn − a is irreducible,

showing that (2) ⇔ (3).


Suppose that (1) is false, so a = bm for some b ∈ F. Let n = jm. ThenXn − a = X jm − bm has X j − b as a proper factor, so it is not irreducible.

On the other hand, suppose that (3) is false. Then, as we have observed,mα(X) = ∏m−1

k=0 (X − ζ ikn α) and {ζ ik } form a group of order m isomorphic

to a subgroup of Z/nZ, so it must be {ζ km | k = 0, . . . , m − 1}, and then

mα(X) = ∏m−1k=0 (X − ζ k

mα) = X − αm = X − b ∈ F[X ]. But if n = mj , thena = αn = αmj = (αm) j = b j , b ∈ F, so α is not n-powerless in F. ��

This proposition has a partial converse.

Proposition 3.7.7. Let F ⊇ F0(ζn). Let E be an extension of F. The followingare equivalent:

(1) E is the splitting field of Xn −a ∈ F[X ], for some a that is n-powerlessin F.

(2) E is a Galois extension of F with Gal(E/F) ∼= Z/nZ.

Proof. (1) implies (2) is (1) implies (2) of Proposition 3.7.6.Now suppose (2) is true. Let σ be a generator of Gal(E/F). By Theorem

3.6.2, E has a normal basis {σ i (α) | i = 0, . . . , n − 1} for some α ∈ E. Let

β = α + ζ−1n σ(α) + · · · + ζ−(n−1)

n σ n−1(α).

Since {σ i (α)} are independent, β �= 0; then direct calculation shows

σ(β) = σ(α) + ζ−1n σ 2(α) + · · · + ζ−(n−1)

n α = ζnβ,

and so σ i (β) = ζ inβ for each i . Thus {σ i (β)} are all distinct, so E = F(β) by

Lemma 3.5.4. Furthermore,

mβ(X) =n−1∏i=0

(X − ζ inβ) = Xn − βn ∈ F[X ].

Thus, if a = βn , a ∈ F[X ] and E is the splitting field of Xn − a. Finally,a is n-powerless in F as in the proof of Proposition 3.7.6: If a = bm , n = mj ,then mβ(X) would have X j − b as a factor, and mβ(X) is irreducible so j = nand m = 1. ��Remark 3.7.8. Consider the case F0 = Q. In order to apply Propositions 3.7.6and 3.7.7, we must do arithmetic in the field Q(ζn), which may not be so easy.It would be a lot simpler if we only had to do arithmetic in Q, which is easy.Fortunately, this is often, and indeed usually, the case. We have the followingresults, which we shall prove later on.


Lemma 4.6.5. (1) Let n be an odd integer, and let a be n-powerlessin Q. Then a is a n-powerless in Q(ζN ) for every N.(2) Let n be an even integer, and let a be n-powerless in Q. If a isnegative, assume also that −a is not a square in Q. Then one of thefollowing two alternatives holds:(a) a is n-powerless in Q(ζN ) for every N.(b) a = b2 for some b ∈ Q(ζN ), for some N, and b is n/2-powerlessin Q(ζN ′) for every multiple N ′ of N.Corollary 4.5.5. Let a be a square-free integer. If a is positive andodd, set a′ = a. Otherwise set a′ = 4|a|. Then a = b2 for someb ∈ Q(ζn) if and only if a′ divides n. �

Definition 3.7.9. Let F ⊇ F0(ζn). A Kummer field E is the splitting field of apolynomial of the form

f (X) = (Xn − a1)(Xn − a2) · · · (Xn − at ), ai ∈ F, i = 1, . . . , t. �Recall that the exponent e of a finite group G is the smallest positive integer

with the property that ge = 1 for every g ∈ G, and that the exponent of a groupdivides its order.

Theorem 3.7.10. Let F ⊇ F0(ζn), and let E be an extension of F. Then E is aKummer field if and only if:

(1) E is a Galois extension of F.(2) Gal(E/F) is abelian.(3) exponent (Gal(E/F)) divides n.

Proof. First suppose that E is a Kummer field, the splitting field of f (X) =(Xn − a1) · · · (Xn − at ). Let αi ∈ E with αn

i = ai for each i . Then Xn − ai =∏nj=1(X − ζ

jn αi ), so Xn − ai splits into a product of distinct linear factors and

is separable; then f (X), the product of separable polynomials, is separable aswell. Thus E is the splitting field of a separable polynomial in F[X ], so E is aGalois extension of F, proving (1). We prove (2) and (3) by induction on t . Thet = 1 case follows immediately from Proposition 3.7.6. Assume the theorem istrue for t −1. Let B1 ⊆ E be the splitting field of f1(X) = (Xn −a1) · · · (Xn −at−1) over F and let B2 ⊆ E be the splitting field of f2(X) = Xn − at overF. Then E = B1B2. Then, by Corollary 3.4.7, Gal(E/F) is a subgroup ofGal(B1/F) × Gal(B2/F), so by the inductive hypothesis (for f1(X)) and thet = 1 case (for f2(X)) Gal(E/F) is a subgroup of an abelian group of exponentdividing n, so Gal(E/F) is itself an abelian group of exponent dividing n, andby induction the result follows.


Conversely, suppose G = Gal(E/F) is as claimed. Write G as a direct sumof cyclic groups G = H1 ⊕ · · · ⊕ Hs . Since the exponent of G divides n, foreach i the order mi of Hi divides n.

We prove the theorem by induction on s. If s = 1, then G = H1 iscyclic of order m, so, by Proposition 3.7.7, E is the splitting field of Xm − b1

for some b1 ∈ F or equivalently (since F ⊇ F0(ζn)) of Xn − a1 wherea1 = bn/m

1 ∈ F. Now suppose the theorem is true for s − 1 and let G bewritten as above. Set B1 = Fix(H1 ⊕ · · · ⊕ Hs−1) and B2 = Fix(Hs). Then,by the inductive hypothesis, B1 is the splitting field of a polynomial g1(X) ofthe form g1(X) = (Xn − a1)(Xn − a2) · · · (Xn − ar ) and, by the s = 1 case,B2 is the splitting field of a polynomial g2(X) of the form g2(X) = Xn −ar+1. Then, by Proposition 2.8.14 (1), B1B2 = Fix((H1 ⊕ · · · ⊕ Hs−1) ∩Hs) = Fix({1}) = E and E = B1B2 is the splitting field of g1(X)g2(X) =(Xn − a1)(Xn − a2) · · · (Xn − ar+1), and by induction the result follows. ��

Let us now explicitly determine the Galois group of a Kummer extension.For a field F, recall that F∗ denotes the group (under multiplication) of nonzeroelements of n. We let (F∗)n denote those elements that are nth powers, andobserve that (F∗)n is a subgroup of F∗.

Theorem 3.7.11. Let F ⊇ F0(ζn) and let E be the splitting field of

f (X) = (Xn − a1) · · · (Xn − at )

with each ai �= 0. Then Gal(E/F) is isomorphic to 〈a1, . . . , at 〉 ⊆ F∗/(F∗)n

(i.e., to the subgroup of F∗/(F∗)n generated by a1, . . . , at ).

Proof. Let αi ∈ E∗ with αni = ai , i = 1, . . . , n. Let G = 〈α1, . . . , αt 〉 ⊂

E∗/F∗ and let G ′ = 〈a1, . . . , at 〉 ⊆ F∗/(F∗)n . Let ϕ : G → G ′ by ϕ(α) = αn .We claim ϕ is an isomorphism. Clearly, ϕ is an epimorphism, so we need onlyshow it is a monomorphism. Suppose ϕ(α) = 1. Then αn ∈ (F∗)n , so αn = f n

for some f ∈ F∗; then α = ζ kn f ∈ F∗ for some k (as ζn ∈ F by hypothesis).

Thus instead of considering G ′ we may consider G.Now write G as a direct sum of cyclic subgroups G = H1 ⊕ · · · ⊕ Hs and

let βi ∈ E be a generator of Hi . Let Hi have order mi dividing n. Set bi = βmii ,

so bi ∈ F and bi is mi -powerless in F. (Then also G ′ = H ′1 ⊕ · · · ⊕ H ′

s withH ′

i the subgroup generated by bi .)We prove the corollary by induction on s. We include in the inductive

hypothesis that Gal(E/F) is generated by elements ηi with ηi (βi ) = ζmi βi andηi (β j ) = β j for j �= i . If s = 1, this hypothesis is just the conclusion ofProposition 3.7.7.

Now assume the hypothesis is true for s − 1. Let B = F(β1, . . . , βs−1) andD = F(βs).


Claim: B and D are disjoint extensions of F with composite BD = E.

Assuming this claim, the inductive step, and hence the corollary then fol-lows immediately from Theorem 3.4.7.

Proof of claim: E = F(α1, . . . , αt ) and 〈α1, . . . , αt 〉 = 〈β1, . . . , βs〉 ⊂E∗/F∗, so certainly BD = E.

Let D0 = B ∩ D. By the FTGT, B0 = Fix(K ) for some subgroup K ofGal(B/F), which is a cyclic group of order ms . From this we see that D0 =F((βs)

m ′) = F( m

√bs), where mm ′ = ms , for some m dividing ms . We want to

show m = 1.Let us set β = m

√bs = (βs)

m ′. Then β ∈ B, so by the inductive hypothesis

we may write uniquely

β = (βs)m ′ =

∑i1,...,is−1

ci1,...,is−1βi11 · · · β is−1

s−1

with i j = 0, . . . , m j − 1, for each j = 1, . . . , s − 1, and with each ci1,...,is−1

∈ F.Now let σ ∈ Gal(B/F). Then σ(β) = ζβ for some root of unity ζ , and,

similarly, σ(βi11 · · · β is−1

s−1) = ζ ′β i11 · · · β is−1

s−1 for some root of unity ζ ′. Thus wesee that we must have ζ ′ = ζ for every term with ci1,...,is−1 �= 0. But, by theinductive hypothesis, we may choose σ with σ(β j ) = ζ jβ j where each ζ j isan arbitrary m th

j root of 1. Hence, if this summation has more than one nonzeroterm, we may find a σ which has the effect of multiplying one term by ζ andanother term by ζ ′ with ζ ′ �= ζ , which is impossible. Hence we can concludethere is only one term in the summation, and that

β = (βs)m ′ = ci1,...,is−1β

i11 · · · β is−1

s−1;

so we see that (βs)m ′ ∈ H1 ⊕ . . . ⊕ Hs−1 ⊂ E∗/F∗. But by assumption G =

H1 ⊕· · ·⊕ Hs ; in particular (H1 ⊕ . . .⊕ Hs−1)∩ Hs = {1}, and hence we musthave m ′ = ms and hence m = 1, as claimed. ��Corollary 3.7.12. In the situation of, and in notation of the proof of Theorem3.7.11, the isomorphism ⊕s

i=1(Z/mi Z) → Gal(E/F) is given by

( j1, . . . , js) �→ σ where σ(βi ) = ζ ji n/min βi ( = ζ ji

miβi ), i = 1, . . . , s.

Proof. This follows directly from the proof of Theorem 3.7.11. ��Example 3.7.13. (1) Let E be the splitting field of (X2 − 2)(X2 − 3)(X2 − 5)

over Q. Then Gal(E/Q) ∼= (Z/2Z)3 as 〈2, 3, 5〉 ⊆ Q∗/(Q∗)2 is isomorphic to(Z/2Z)3.

3.8 The Norm and Trace 79

(2) Let E be the splitting field of (X2 − 2)(X2 − 3)(X2 − 6) over Q. ThenGal(E/Q) ∼= (Z/2Z)2 as 〈2, 3, 6〉 ⊆ Q∗/(Q∗)2 is isomorphic to (Z/2Z)2.

(3) Let E be the splitting field of X12 −5 over Q(ζ12). Then Gal(E/Q(ζ12))∼= Z/12Z as 〈5〉 ⊆ Q(ζ12)∗/(Q(ζ12)

∗)12 is isomorphic to Z/12Z. The polyno-mial X12 − 5 is irreducible over Q and also over Q(ζ12).

(4) Let E be the splitting field of X12 −3 over Q(ζ12). Then Gal(E/Q(ζ12))∼= Z/6Z as 〈3〉 ⊆ Q(ζ ∗12)/(Q(ζ12)

∗)12 is isomorphic to Z/6Z. Note√

3 ∈Q(ζ12). The polynomial X12 − 3 is irreducible over Q. Over Q(ζ12), it factorsas the product of two irreducible factors (X6 − √

3)(X6 + √3).

(5) Let E be the splitting field of X4 + 1 over Q(ζ4). Then E = Q(ζ8),Gal(E/Q(ζ4)) ∼= Z/2Z, and 〈−1〉 ⊆ Q(ζ4)

∗/(Q(ζ4)∗)4 ∼= Z/2Z as −1 is a

square but not a 4th power in Q(ζ4). The polynomial X4 + 1 is irreducibleover Q. Over Q(ζ4), it factors as the product of two irreducible polynomials(X2 + i)(X2 − i).

(6) Let E be the splitting field of X4 + 4 over Q(ζ4). In fact E = Q(ζ4) asX4 + 4 = (X − (1 + i))(X − (1 − i))(X − (−1 + i))(X − (−1 − i)); thusGal(E/Q) = {id}, and 〈−4〉 ⊆ Q(ζ4)

∗/(Q(ζ4)∗)4 ∼= {id} as −4 = (1 + i)4, an

equation in Q(ζ4).In this example we have made a number of claims about arithmetic in

cyclotomic fields (i.e., in fields Q(ζn)). They will be verified in Section 4.2. �

3.8 The Norm and Trace

Let E be a finite extension of F. We shall define two important functions fromE to F, the norm and the trace. We shall begin by defining them in case E is aGalois extension of F, which is the most important case, and the case to whichwe denote most of our attention. At the end, we will show how to generalizethe definition.

Definition 3.8.1. Let E be a finite Galois extension of F with Galois groupG = Gal(E/F). For α ∈ E, the norm (from E to F) of α is

NE/F(α) =∏σ∈G

σ(α)

and the trace (from E to F) of α is

TrE/F(α) =∑σ∈G

σ(α). �

Lemma 3.8.2. For any α ∈ E, NE/F(α) and TrE/F(α) are elements of F.


Proof. NE/F(α) and TrE/F(α) are obviously invariant under G. ��Lemma 3.8.3. Let α ∈ E with deg mα(X) = r , and write mα(X) =∑r

i=0 ci Xi . Let (E/F) = n. Then

NE/F(α) = (−1)ncn/r0 and TrE/F(α) = −(n/r)cr−1.

Proof. If α has distinct conjugates α1, . . . , αr (with α1 = α), then mα(X) =∏ri=1(X − αi ) by Lemma 2.7.12, so

c0 = (−1)rα1 · · · αr and cr−1 = −(α1 + · · · + αr ). ��Lemma 3.8.4. For α ∈ E, let Tα : E → E be the linear transformation ofF-vector spaces given by Tα(β) = αβ. Then

NE/F(α) = Det(Tα) and TrE/F(α) = Trace(Tα).

Proof. Let B = F(α). Then B has F-basis B = {1, α, . . . , αr−1} wherer = deg mα(X). If (E/F) = n, then (E/B) = n/r . Choose a B-basis{1 = ε1, . . . , εn/r } for E over B.

Then

C = {1, α, . . . , αr−1, . . . , εn/r , εn/rα, . . . , εn/rαr−1}

is a basis for E over F.Let Sα = Tα | F(α). Now Sα : F(α) → F(α) and the matrix of Sα in

the basis B is just C(mα(X)), the companion matrix of the polynomial mα(X)

(see the proof of Proposition 2.4.6(3)). Then the matrix of Tα in the basis Cis a block diagonal matrix consisting of n/r diagonal blocks, each equal toC(mα(X)), and the result then follows from Lemma 3.8.3. ��Theorem 3.8.5. Let E be a finite Galois extension of F of degree n. Let B bea field intermediate between F and E. (Then E is a Galois extension of B.)Assume further that B is a Galois extension of F. Then for any element a of Fand any elements α, β of E:

(1) NE/F(a) = an and TrE/F(a) = na.(2) NE/F(αβ) = NE/F(α) NE/F(β) and TrE/F(α + β) = TrE/F(α) +

TrE/F(β).(3) NE/F(α) = NB/F(NE/B(α)) and TrE/F(α) = TrB/F(TrE/B(α)).

Proof. (1) and (2) are clear. We prove (3). Let H = Gal(E/B) = {τi }have left coset representatives {ρ j }. Then we may identify G/H with {ρ j }.Then NB/F(NE/B(α)) = ∏

j ρ j (∏

i τi (α)) = ∏i, j ρ jτi (α) = ∏

σ∈G σ(α) =NE/F(α), and similarly for TrE/F(α). ��

3.8 The Norm and Trace 81

Lemma 3.8.6. There is an element α ∈ E with TrE/F(α) �= 0.

Proof. If char(F) = 0 or n = (E/F) is relatively prime to p = char(F), this iscompletely trivial, as we may simply choose α = 1.

In any case, for each σ ∈ Gal(E/F), σ : E∗ → E∗ is a character ofthe group E∗ in E in the sense of Definition 2.8.1, so the result follows im-mediately from Dirichlet’s theorem on the independence of characters (Theo-rem 2.8.4). ��Theorem 3.8.7 (Hilbert’s Theorem 90). Let E be a cyclic extension of F ofdegree n and let σ be a generator of G = Gal(E/F).

(1) If β is any element of E with NE/F(β) = 1, then β = α/σ(α) for someα ∈ E.

(2) If β is any element of E with TrE/F(β) = 0, then β = α − σ(α) forsome α ∈ E.

Proof. (1) By Theorem 2.8.4, {1, σ, . . . , σ n−1} are independent characters ofE∗ in E, so for any elements ε0, . . . , εn−1 of E, there is an element δ of E withα = ∑n−1

i=0 εiσi (δ) �= 0. Now let

εi = σ o(β)σ 1(β) · · · σ i−1(β), i = 1, . . . , n − 1.

Note that βσ(εi ) = εi+1 for i = 1, . . . , n − 2, and εn−1 = NE/F(β) = 1so also βσ(εn−1) = β = σ o(β) = ε1. Let α be as above and note then thatβσ(α) = α so β = α/σ(α).

(2) By Lemma 3.8.6, there is an element of γ of E with TrE/F(γ ) �= 0. Setδ = βσ(γ ) + (β + σ(β))σ 2(γ ) + · · · + (β + · · · + σ n−2(β))σ n−1(γ ).

Then, by direct calculation,

δ − σ(γ ) = β(σ(γ ) + · · · + σ n−1(γ )) − (σ (β) + · · · + σ n−1(β))γ

= β TrE/F(γ ) − γ TrE/F(β) = βT rE/F(γ ),

so if α = δ/ TrE/F(γ ), we have that β = α − σ(α). ��Now we extend the definitions of norm and trace. First we consider sep-

arable extensions. Here we assume familiarity with Section 5.2 below.

Definition 3.8.8. Let E be a finite separable extension of F. Then

NE/F = ND/F and TrE/F = TrD/F

where D is any normal closure of E. �


Remark 3.8.9. (1) It is easy to see that NE/F and TrE/F are well defined, i.e.,independent of the choice of D.

(2) The results of this section through Theorem 3.8.5 go through un-changed in this more general situation. �

Now we consider arbitrary finite extensions. Here we further assume fa-miliarity with Section 5.1 below.

Definition 3.8.10. Let E be a finite extension of F. Then

NE/F = (ND/F)d and Tr E/F = d TrD/F,

where D is any normal closure of E and d = (Fix(Gal(D/F))/F). �Remark 3.8.11. (1) Let Fi = Fix(Gal(D/F)). Then D is a Galois extension ofFi and Fi is a purely inseparable extension of F. If E is a separable extensionof F, then D is a Galois extension of F, Fi = F, and d = 1. In particular thisis always the case for a perfect field F.

(2) If char(F) = p and E is not a separable extension of F, then d is apower of p and so in this case TrE/F = 0.

(3) Again the results of this section through Theorem 3.8.5 continue tohold. �

3.9 Exercises

Exercise 3.9.1. Give an example of an extension E of F of degree n and adivisor d of n for which there does not exist a field B intermediate between Fand E with (B/F) = d.

Exercise 3.9.2. Let F be a field with char(F) �= 2. Call a rational func-tion f (X1, . . . , Xn) ∈ F(X1, . . . , Xn) alternating if f (σ (X1), . . . , σ (Xn)) =sign(σ ) f (X1, . . . , Xn) for every σ ∈ Sn . Let s1, . . . , sn denote (as usual) theelementary symmetric functions in X1, . . . , Xn . Let δ be as in Definition 4.8.1.(a) Show that every alternating rational function f (X1, . . . , Xn) can be writtenas a rational function in s1, . . . , sn and δ in a unique way.(b) Show that in fact every alternating rational function f (X1, . . . , Xn) canbe written as f (X1, . . . , Xn) = g(s1, . . . , sn)δ for some rational functiong(s1, . . . , sn) of s1, . . . , sn .(c) Show that every alternating polynomial f (X1, . . . , Xn) can be written as apolynomial in s1, . . . , sn and δ in a unique way.(d) Show that in fact every alternating polynomial f (X1, . . . , Xn) can be

3.9 Exercises 83

written as f (X1, . . . , Xn) = g(s1, . . . , sn)δ for some polynomial functiong(s1, . . . , sn) in s1, . . . , sn .(e) Show that every rational function f (X1, . . . , Xn) invariant under the ac-tion of An (i.e., every rational function f (X1, . . . , Xn) with f (σ (X1), . . . ,

σ (Xn)) = f (X1, . . . , Xn) for every σ ∈ An) can be written in a unique way asf (X1, . . . , Xn) = fs(X1, . . . , Xn) + fa(X1, . . . , Xn) where fs(X1, . . . , Xn)

is symmetric and fa(X1, . . . , Xn) is alternating. Furthermore, if f (X1, . . . ,

Xn) is a polynomial, then so are fs(X1, . . . , Xn) and fa(X1, . . . , Xn).

Exercise 3.9.3. Show that (2) implies (1) in Exercise 2.10.6.

Exercise 3.9.4. Show that (2) implies (1) in Exercise 2.10.7.

Exercise 3.9.5. Interpret α in Exercise 2.10.1 as being in an extension of F5,find an irreducible polynomial in F5[X ] that has α as a root, and find all rootsof that polynomial in a splitting field E. Also, find G = Gal(E/F). (In thedefinition of α, there may be an ambiguity because of the roots. Handle allcases.)

Exercise 3.9.6. Interpret α in Exercise 2.10.1 as being in an extension of F7,find an irreducible polynomial in F7[X ] that has α as a root, and find all rootsof that polynomial in a splitting field E. Also, find G = Gal(E/F). (In thedefinition of α, there may be an ambiguity because of the roots. Handle allcases.)

Exercise 3.9.7. In this exercise, let E be a splitting field of the polynomialf (X) ∈ F[X ] and let G = Gal(E/F). Find (E/F) and find G.

(a) f (X) = X3 − 1 ∈ F3[X ].(b) f (X) = X3 − 1 ∈ F5[X ].(c) f (X) = X3 − 1 ∈ F7[X ].(d) f (X) = X3 − 3 ∈ F5[X ].(e) f (X) = X3 + 4X + 2 ∈ F5[X ].(f) f (X) = X4 + 2X3 + 3X2 + 3X + 4 ∈ F5[X ].(g) f (X) = X4 + 3X2 + 4 ∈ F5[X ].(h) f (X) = X4 + 3X2 + 3 ∈ F5[X ].

Exercise 3.9.8. Let F = Fp and let f (X) = X p − X − a for some a �= 0 ∈ F.Let E be the splitting field of F and let α be a root of f (X) in E. Show thatα, α + 1, . . . , α + (p − 1) are the roots of f (X) in E. Conclude that f (X) ∈F[X ] is irreducible and that Gal(E/F) is cyclic of order p with generator σ theautomorphism of E given by σ(α) = α+1. Then the Frobenius automorphism of F must be given by (α) = α + k for some k �= 0 ∈ F. Show that acan be chosen so that (α) = α + 1. (E is called an Artin–Schreier extensionof F.)


Exercise 3.9.9. Let f (X) ∈ Z[X ] be a monic polynomial and let f (X) beits image in Fp[X ] (identifying Fp with Z/pZ). Let E be a splitting field forf (X) and let E be a splitting field for f (X). Show that if, for some p, all theroots of f (X) in E are simple, then all the roots of f (X) in E are simple.

Exercise 3.9.10. Let B1 and B2 be Galois extensions of F, both contained insome field A. Let E be the composite E = B1B2.(a) Show that every σ ∈ Gal(B1/F) extends to a (not necessarily unique)automorphism σ ∈ Gal(E/F).(b) Show that B1 and B2 are disjoint extensions of F if and only if every σ ∈Gal(B1/F) extends to an automorphism σ ∈ Gal(E/F) with σ | B2 = id.

Exercise 3.9.11. Find an example of a field F and pairwise disjoint Galoisextensions B0, B1, B2, of F with B0 and B1B2 not disjoint extensions of F.

Exercise 3.9.12. Let D be a Galois extension of F and let B be an extension ofF. Suppose that D and B are disjoint extensions of F.(a) Show that D and B′ are disjoint extensions of F for any Galois conjugateB′ of B.(b) Let B be the Galois closure of B. Give an example to show that D and Bneed not be disjoint.

Exercise 3.9.13. Let B1 and B2 be finite Galois extensions of F. Show thatB1 ∩ B2 is a Galois extension of F.

Exercise 3.9.14. (a) Let a ∈ F and suppose that a is not a pth power in F. LetB be an extension of F of degree m, and suppose that m and p are relativelyprime. Show that a is not a pth power in B.(b) More generally, let a ∈ F and suppose that a is n-powerless in F. Let B bean extension of F of degree m, and suppose that m and n are relatively prime.Show that a is n-powerless in B.

Exercise 3.9.15. Suppose that Xm − a and Xn − a are both irreducible poly-nomials in F[X ].(a) Suppose that m and n are relatively prime. Show that Xmn − a is an irre-ducible polynomial in F[X ].(b) More generally, let � = lcm(m, n). Show that X � − a is an irreduciblepolynomial in F[X ].Exercise 3.9.16. (a) Let E = Q(

√2,

√3,

√5,

√7, . . . ). Show that E/Q is

infinite.(b) Let E = Q(

√2,

3√

2,4√

2,5√

2, . . . ). Show that E/Q is infinite.

3.9 Exercises 85

Exercise 3.9.17. (a) Let F ⊇ F0(ζ2) (and consequently char(F) �= 2) and letf (X) = X2 − a ∈ F[X ] be irreducible. Let E = F(α) where α2 = a. Letβ = b0 + b1α with b0, b1 ∈ F. Show that the conjugates of β form a normalbasis for E if and only if b0b1 �= 0.(b) Let F ⊇ F0(ζ3) (and consequently char(F) �= 3) and let f (X) = X3 − a ∈F[X ] be irreducible. Let E = F(α) where α3 = a. Let β = b0 + b1α + b2α

2

with b0, b1, b2 ∈ F. Show that the conjugates of β form a normal basis for Eif and only if b0b1b2 �= 0.

Exercise 3.9.18. Let F ⊆ B ⊆ E with E/F Galois. Show that there existirreducible polynomials f (X) ∈ F[X ] and g(X) ∈ B[X ] such that E is asplitting field of f (X), and E is also a splitting field of g(X), and g(X) dividesf (X) in B[X ].Exercise 3.9.19. (a) Let E be a separable extension of F and suppose there isan integer n such that deg mα(X) ≤ n for every α ∈ E. Show that E is a finiteextension of F. Furthermore, show that (E/F) ≤ n.(b) Give an explicit counterexample to (a) in case E is not a separable extensionof F.

Exercise 3.9.20. Let E be an extension of F of degree 2, where F = Fq isa finite field, with q odd. Let E = F(α), where α2 ∈ F. Show that, for anya, b ∈ F,

(a + bα)q = a − bα or, equivalently, NE/F(α) = (a + bα)q+1.

Exercise 3.9.21. Let E be an extension of F of degree 2, where char(F) �= 2.Let E = F(δ), where δ2 = d ∈ F. Let T : E → E be an F-linear transforma-tion. Let α = T (1) and β = T (δ)/δ. (Note α, β ∈ E.)(a) If α = ±β, show that there exists an f ∈ F with

NE/F(T (ε)) = f NE/F(ε) for all ε ∈ E

and in fact f = NE/F(α) = NE/F(β).(b) If α �= ±β, show there does not exist such an f .

Exercise 3.9.22. Let E be an extension of F of degree 2, where char(F) �= 2.Let E = F(δ), where δ2 = d ∈ F, and let σ be the nontrivial element ofGal(E/F). Prove Hilbert’s Theorem 90 (Theorem 3.8.7 (1)) directly in thiscase. That is, if γ = a + bδ ∈ E with NE/F(γ ) = a2 − b2d = 1, then there isan element θ = x + yδ ∈ E with

γ = θ

σ (θ)= x + yδ

x − yδ.


Exercise 3.9.23. Let F = Fps and let E = Fpr for some multiple r of s. Setm = (pr − 1)/(ps − 1). Recall that Gal(E/F) is cyclic of order r/s withgenerator � given by �(α) = α ps

.(a) Show that NE/F(α) = αm for every α ∈ E.(b) Show that NE/F : E → F is an epimorphism.(c) Prove Hilbert’s Theorem 90 (Theorem 3.8.7 (1)) directly in this case: Ifβ ∈ E with NE/F(β) = 1, then β = (α)(α−ps

) for some α ∈ E.

Exercise 3.9.24. (a) Fix a polynomial k(Y ) ∈ F[Y ]. Let f (X) ∈ F[X ] beany separable polynomial. Let f (X) split in E with roots α1, . . . , αm . LetQk(Y )( f (X)) be the polynomial

Qk(Y )( f (X)) =∏

i

(X − k(αi )).

Evidently Qk(Y )( f (X)) ∈ E[X ]. Show that in fact Qk(Y )( f (X)) ∈ F[X ].(b) Fix a polynomial k(Y, Z) ∈ F[Y, Z ]. Let f (X), g(X) ∈ F[X ] be anyseparable polynomials. Let f (X) and g(X) split in E with roots α1, . . . , αm

and β1, . . . , βn , respectively. Let Qk(Y,Z)( f (X), g(X)) be the polynomial

Qk(Y,Z)( f (X), g(X)) =∏i, j

(X − k(αi , β j )).

Evidently Qk(Y,Z)( f (X), g(X)) ∈ E[X ]. Show that in fact

Qk(Y,Z)( f (X), g(X)) ∈ F[X ].(This generalizes to arbitrarily many polynomials.)

Exercise 3.9.25. (a) Let {si } be the elementary symmetric polynomials in {αi }.Show that for any k(Y ) ∈ F[Y ], Qk(Y )( f (X)) is a polynomial whose coeffi-cients are polynomials in {si }, and hence polynomials in the coefficients off (X).(b) Let {si } be the elementary symmetric polynomials in {αi } and let {s ′

j } bethe elementary symmetric polynomials in {β j }. Show that for any k(Y, Z) ∈F[Y, Z ], Qk(Y,Z)( f (X), g(X)) is a polynomial whose coefficients are polyno-mials in {si } and {s ′

j }, and hence polynomials in the coefficients of f (X) andg(X). (This generalizes to arbitrarily many polynomials.)

Exercise 3.9.26. (a) Show that if k(Y ) ∈ Z[Y ] and f (X) ∈ Z[X ], thenQk(Y )( f (X)) ∈ Z[X ].(b) Show that if k(Y, Z) ∈ Z[Y, Z ] and f (X), g(X) ∈ Z[X ], thenQk(Y,Z)( f (X), g(X)) ∈ Z[X ]. (This generalizes to arbitrarily many polyno-mials.)

3.9 Exercises 87

Exercise 3.9.27. Let f (X) = X2 + aX + b, g(X) = X2 + cX + d, andh(X) = X3 + eX2 + f X + g be polynomials in F[X ]. Without finding theroots of these polynomials:(a) Find QY+1( f (X)), Q2Y ( f (X)), QY 2( f (X)), and QY 3( f (X)).(b) Find QY+1(h(X)), Q2Y (h(X)), and QY 2(h(X)).(c) Find QY+Z ( f (X), g(X)) and QY Z ( f (X), g(X)).(d) Find QY+Z ( f (X), h(X)) and QY Z ( f (X), h(X)).

Exercise 3.9.28. Let R be a subring of F and let E be an extension of F. LetS = {α ∈ E | α is integral over R}. Show that S is a subring of E.

Exercise 3.9.29. Let E be a field and let R ⊆ E. Suppose that E = R[α] forsome α ∈ E. Show that there is a nonzero element s ∈ R such that R[1/s] is afield.

4

Extensions of the Field of Rational Numbers

In this chapter, we assume that all fields we consider are of characteristic 0.

4.1 Polynomials in Q[X]In this section we deal with a number of questions about polynomials in Q[X ]related to factorization and irreducibility.

We begin with a couple of elementary results.

Lemma 4.1.1. Let F ⊆ R and let f (X) ∈ F[X ]. If γ ∈ C is a root of f (X),then its complex conjugate γ is also a root of f (X). Consequently, if f (X) hasodd degree, then it has an odd number of real roots, counted with multiplicity,and if f (X) has even degree, then it has an even number of real roots, countedwith multiplicity.

Proof. f (γ ) = f (γ ) = f (γ ) = 0 = 0. ��Lemma 4.1.2. Let f (X) ∈ Z[X ], f (X) = ∑n

i=0 ai Xi . Suppose that r ∈ Q isa root of f (X) (or equivalently, that X − r is a factor of f (X)). If r = s/t ,with s and t relatively prime, then s divides a0 and t divides an. In particular,if f (X) is monic, then r is an integer dividing a0.

Proof. 0 = f (s/t) = tn f (s/t) = ∑ni=1 ai si tn−i . Now 0 is divisible by s, and

every term of this summation except possibly a0tn is divisible by s, so thatterm must be divisible by s as well, and since t is relatively prime to s, thisimplies that a0 is divisible by s. By the same logic applied to the term ansn , an

is divisible by t . ��


90 4 Extensions of the Field of Rational Numbers

Note that this lemma, used recursively, provides an effective method forfinding all linear factors of f (X) in Q[X ].

Corollary 4.1.6 below is a standard result, but one we shall use repeatedly.First, some preliminaries.

Definition 4.1.3. A polynomial f (X) = an Xn + · · · + a0 ∈ Z[X ] is primitiveif gcd(an, . . . , a0) = 1. �Lemma 4.1.4 (Gauss’s Lemma). If g(X) and h(X) are primitive polynomialsin Z[X ], and f (X) = g(X)h(X), then f (X) is a primitive polynomial inZ[X ].Proof. Write g(X) = bm Xm + · · · + b0 and h(X) = ck Xk + · · · + c0. Letf (X) = g(X)h(X) = an Xn + · · · + a0. Suppose f (X) is not primitive andlet d = gcd(an, . . . , a0). Pick a prime p dividing d. By hypothesis, p does notdivide all of the coefficients of g(X), so let i be the smallest index with bi notdivisible by p. Similarly, p does not divide all of the coefficients of h(X), solet j be the smallest index with c j not divisible by p.

Consider ai+ j , the coefficient of Xi+ j in f (X). Then

ai+ j = bi c j + (bi+1c j−1 + bi+2c j−2 + · · · )+ (bi−1c j+1 + bi−2c j+2 + · · · ).

By hypothesis, p divides ai+ j , and by construction p divides each of the termsin the two parenthesized expressions, so p divides bi c j as well, and hence pdivides bi or p divides c j , a contradiction. ��Proposition 4.1.5. Let f (X) be a polynomial in Z[X ] and suppose f (X) =g(X)h(X) with g(X), h(X) ∈ Q[X ]. Then f (X) = g1(X)h1(X) with g1(X),h1(X) ∈ Z[X ], where g1(X) is a constant multiple of g(X), and h1(X) is aconstant multiple of h(X).

Proof. We may write f (X) = e f0(X) where f0(X) ∈ Z[X ] is primitive ande ∈ Z. We may write g(X) = cg0(X) where g0(X) ∈ Z[X ] is primitive andc ∈ Q. Similarly we may write h(X) = dh0(X) where h0(X) ∈ Z[X ] is prim-itive and d ∈ Q. Then e f0(X) = f (X) = g(X)h(X) = cdg0(X)h0(X).By Lemma 4.1.4, g0(X)h0(X) is primitive so we must have e = ±cd,so f0(X) = ±g0(X)h0(X) and f (X) = g1(X)h1(X) where g1(X) =±eg0(X) = (±e/c)g(X) and h1(X) = h0(X) = (1/d)h(X). ��Corollary 4.1.6. Let f (X) be a monic polynomial in Z[X ] and supposef (X) = g(X)h(X) with g(X), h(X) ∈ Q[X ] monic polynomials. Theng(X), h(X) ∈ Z[X ]. Consequently, if f (X) is irreducible in Z[X ], then f (X)

is irreducible in Q[X ].

4.1 Polynomials in Q[X ] 91

Proof. Apply Proposition 4.1.5, and note that we must have g1(X) = g(X)

and h1(X) = h(X) (up to a common sign) as f (X), g(X), and h(X) aremonic. The second claim follows by contraposition. ��

We have the following useful criterion for a polynomial to be irreducibleover Q.

Proposition 4.1.7 (Eisenstein’s Criterion). Let f (X) be a polynomial withinteger coefficients, f (X) = an Xn + an−1 Xn−1 + · · · + a0. Suppose there is aprime p with p not dividing an, p dividing a0, . . . , an−1, and p2 not dividinga0. Then f (X) is irreducible in Q[X ].Proof. By Corollary 4.1.6, it suffices to show that f (X) is irreducible in Z[X ].Suppose f (X) = g(X)h(X) with g(X) = bm Xm + bm−1 Xm−1 + · · · + b0 andh(X) = ck Xk +ck−1 Xk−1+· · ·+c0. Then bmck = an , so neither is divisible byp. Since p divides a0 = b0c0 but p2 does not divide a0, by hypothesis, p mustdivide exactly one of b0 and c0. Let that one be b0. Now a1 = b1c0+b0c1. Sincep divides a1 and b0 but not c0, p must divide b1. Then a2 = b2c0 +b1c1 +b0c2,so by similar logic p must divide b2. Proceeding inductively, we find that pdivides bm , contradicting bm = 1. ��Example 4.1.8. We will use Eisenstein’s Criterion to show that for a primep the polynomial p(X) = X p−1 + · · · + X + 1 = (X p − 1)/(X − 1)

is irreducible. Of course, Eisenstein’s Criterion does not apply directly. Butobserve that p(X) is irreducible if and only if p(X + 1) is irreducible,p(X + 1) = ((X + 1)p − 1)/X . Expanding by the Binomial Theorem, wesee that p(X + 1) = ∑p

i=1

(pi

)Xi−1 is a monic polynomial of degree p − 1

with the coefficient of every power of X other than X p−1 divisible by p, andwith constant term p divisible by p but not p2, so is irreducible. (p(X) isthe pth cyclotomic polynomial. For n composite, n(X) is not defined by theabove formula. We deal with cyclotomic polynomials in Section 4.2.) �

Let a be an integer of the form a = pq, where p is a prime and p andq are relatively prime. Then, by Eisenstein’s Criterion (Proposition 4.1.7), thepolynomial Xn −a is irreducible over Q for any n. Thus, for example, X3−2 isirreducible over Q. However, Eisenstein’s Criterion does not apply to X3 − 4,which is also irreducible. Let us develop a criterion that does.

Lemma 4.1.9. Let a be an integer of the form a = pmq, where p is a primeand p and q are relatively prime. Let n be any integer that is relatively primeto m. Then the polynomial Xn − a is irreducible over Q.


Proof. Let α = n√

a and E = Q(α). Then (E/Q) = deg mα(X) by Proposition2.4.6 (2). Thus (E/Q) = n if and only deg mα(X) = n. Now mα(X) dividesXn − a (as α is a root of this polynomial), so deg mα(X) = n if and only ifXn − a is irreducible. Thus, putting these together, we see that (E/Q) = n ifand only if Xn − a is irreducible.

Let j be an integer with jm ≡ 1(mod n), so jm = nk + 1 for some k.Clearly Q(α) ⊇ Q(α j ) (and in fact they are equal). Now (α j )n = αnj =a j = pmj q j = pnk+1q j so β = α j/pk satisfies βn = pq j . In other words,β is a root of the polynomial Xn − pq j , and this polynomial is irreducible byEisenstein’s Criterion. Thus (Q(α j )/Q) = (Q(β)/Q) = n so (E/Q) = n asrequired. ��Corollary 4.1.10. (1) Let p be an odd prime. If a ∈ Z is not a pth power, thenX pt − a is irreducible for every t ≥ 0.

(2) If a ∈ Z is positive and not a square, then X2t − a is irreducible forevery t ≥ 0.

Proof. In these cases a satisfies the hypotheses of Lemma 4.1.9 (changingnotation so that n = p in case (1) and n = 2 in case (2).) ��Remark 4.1.11. In the excluded case, X2t − a may not be irreducible. For ex-ample, X4 + 4 = (X2 + 2X + 2)(X2 − 2X + 2). �

Lemma 4.1.9 and Corollary 4.1.10 do not apply to the polynomialX6 − 72. This polynomial is also irreducible, but it will require considerablymore work to show this. We do this in Section 4.6, where we show that Xn −ais irreducible in considerable generality.

Remark 4.1.12. There is an algorithm, due to Kronecker, for factoring polyno-mials f (X) ∈ Q[X ].

It clearly suffices to consider the case that f (X) ∈ Z[X ] is a primitivepolynomial. Then, by Corollary 4.1.6, it suffices to factor f (X) in Z[X ]. Letf (X) have degree n.

We begin by recalling the Lagrange Interpolation Formula. If a1, . . . , am

are distinct, and b1, . . . , bm are arbitrary, there is a unique polynomial g(X) ofdegree at most m − 1 with g(ai ) = bi for each i , given by

g(X) =m∑

i=1

bi

∏j �=i

(X − a j )

(ai − a j ).

Using this we proceed as follows, looking for a factor g(X) of f (X). Ifthere is no such factor, then f (X) is irreducible. If there is such a factor, write

4.2 Cyclotomic Fields 93

f (X) = g(X) f1(X) and proceed recursively (next factoring f1(X), . . . ). Tofind such a factor, let m = 2, . . . , [m/2]+ 1. Pick distinct integers a1, . . . , am .If f (ai ) = 0 for some i , then we have found a linear factor X − ai . Assumenot and let Bi = {bi } be the finite set of divisors of f (ai ), for each i . For eachchoice of elements bi ∈ Bi , let g(X) be the polynomial given by the LagrangeInterpolation Formula with g(ai ) = bi . If g(X) has degree less than m − 1,or does not have integer coefficients, there is nothing to check. If g(X) is apolynomial of degree m − 1 with integer coefficients, test whether f (X) isdivisible by g(X). �Remark 4.1.13. For future reference, we record an extension of this algorithm.Let R be the polynomial ring R = Z[X1, . . . , Xn] and let F be its quotientfield, the field of rational functions in X1, . . . , Xn with coefficients in Q. Letf (Xn+1) ∈ F[Xn+1] be a polynomial. Then there is an algorithm for factoringf (Xn+1).

We proceed by induction. For n = 0 this is just the algorithm of Remark4.1.12. Assume we have an algorithm for n = k and consider n = k + 1.Now R is a UFD (a Unique Factorization Domain), so the analogs of Gauss’sLemma (Lemma 4.1.4), Proposition 4.1.5, and Corollary 4.1.6 continue to holdfor R. Thus we may, as in Remark 4.1.12, assume that f (Xn+1) is a primitivepolynomial in R[Xn+1]. Then proceed as in Remark 4.1.12. The only thing weneed to see is that there is an algorithm for finding each of the sets Bi . But thatfollows by induction, since in factoring f (ai ), we are in the case n = k. �

4.2 Cyclotomic Fields

In this section we study the splitting fields of the polynomials Xn − 1 over Q.Recall that ζn is a primitive nth root of 1 in a field E if ζ n

n = 1 but ζ mn �= 1

for any 1 ≤ m < n. If E is the splitting field of Xn −1 over Q, then E = Q(ζn).In this case, we may simply take ζn = exp(2π i/n), and we do this henceforth.

Definition 4.2.1. n(X) = ∏(X − ζ ) where the product is taken over all

primitive nth roots of 1.n(X) is the nth cyclotomic polynomial and E = Q(ζn) is the nth cyclo-

tomic field. �Remark 4.2.2. Observe that n(X) and Xn − 1 = ∏n−1

i=0 (X − ζ in) are both

products of distinct linear factors in E[X ], and that n(X) divides Xn − 1 inE[X ]. �

We let ϕ(n) (as usual) denote the Euler totient function, ϕ(n) being thenumber of integers between 1 and n that are relatively prime to n.


Lemma 4.2.3. deg n(X) = ϕ(n).

Proof. We observe that ζ is a primitive nth root of 1 if and only if ζ = ζ kn for

some k relatively prime to n, so {primitive nth roots of 1} = {ζ k | 1 ≤ k ≤n, (k, n) = 1} and the cardinality of this set, which by Definition 4.2.1 is thedegree of mζ (X), is ϕ(n). ��Lemma 4.2.4. (1) n(X) ∈ Q[X ] for every n.

(2) If n1 �= n2, then n1(X) and n2(X) are relatively prime in Q[X ].(3) Xn − 1 = ∏

d|n d(X).

Proof. (1) Let E = Q(ζn). By Remark 4.2.2, E is the splitting field of theseparable polynomial Xn−1. Thus, by Theorem 2.7.14, E is a Galois extensionof Q. If ζ is a primitive nth root of 1 in E, then σ(ζ ) is also a primitive nth rootof 1 for any σ ∈ Gal(E/Q), so n(X) is invariant under Gal(E/Q), and hencen(X) ∈ Q[X ].

(2) Let E = Q(ζn1n2). From Definition 4.2.1 we see that n1(X) andn2(X) are relatively prime in E[X ], and hence also in Q[X ] by Lemma 2.2.7.

(3) Every nth root of 1 is a primitive d th root of 1 for a unique ddividing n. ��Corollary 4.2.5. n(X) ∈ Z[X ].Proof. From Lemma 4.1.4 we see that n(X) divides Xn − 1 in Q[X ], andhence, since n(X) is monic, n(X) ∈ Z[X ] by Corollary 4.1.6. ��Theorem 4.2.6. n(X) is irreducible for every n.

First Proof. Suppose n(X) is not irreducible, and consider a monic irre-ducible factor f (X) of n(X) that has ζn as a root. Then f (X) does nothave every primitive nth root of unity as a root, so there is some k > 1with f (ζ k

n ) �= 0. Choose the smallest such k, and let p be any prime factorof k. Since k is relatively prime to n, p is relatively prime to n as well. Setζ = ζ

k/pn . Then, by the minimality of k, f (ζ ) = 0 but f (ζ p) �= 0. Now

f (X) is irreducible and monic, and f (X) divides Xn − 1 in Q[X ], so in factXn − 1 = f (X)g(X) where both f (X) and g(X) are in Z[X ], by Corollary4.1.6. Then

0 = (ζ p)n − 1 = f (ζ p)g(ζ p) and f (ζ p) �= 0, so g(ζ p) = 0.

Writing g(X) = ∑ci Xi , ci ∈ Z, we see that 0 = ∑

ciζpi = ∑

ci (ζp)i , so

h(ζ ) = 0 where h(X) = g(X p). Since f (ζ ) = 0 and f (X) is irreducible

4.2 Cyclotomic Fields 95

and monic, f (X) divides h(X) in Q[X ] and hence in Z[X ], again by Corol-lary 4.1.6. Now let π : Z[X ] → (Z/pZ)[X ] be the map induced by reduc-ing the coefficients mod p. Then π( f (X)) divides π(h(X)) = π(g(X p)) =π(g(X))p in (Z/pZ)[X ], so π( f (X)) and π(g(X)) have an irreducible factorq(X) in common. But

Xn − 1 = π(Xn − 1) = π( f (X))π(g(X)).

This would imply that Xn − 1 is divisible by q(X)2 in (Z/pZ)[X ], and hencethat Xn − 1 has a repeated linear factor in Ep, its splitting field over Z/pZ =Fp. However, Xn − 1 has distinct roots in Ep, a contradiction. (This followsimmediately from Lemma 3.2.1, but is easy to see directly.)

Second Proof (Landau). Let mζn (X) have degree d. Since mζn (X) is monic anddivides Xn −1 in Q[X ], mζn (X) ∈ Z[X ], by Corollary 4.1.6. We claim that, forany k, ζ k

n = rk(ζ ) for a unique polynomial rk(X) ∈ Z[X ] of degree less thand . To see this, note that, by the division algorithm, Xk = mζn (X)qk(X)+rk(X)

with qk(X), rk(X) ∈ Z[X ] unique polynomials with deg rk(X) < d, and thenset X = ζn . It follows immediately that for any polynomial h(X) ∈ Z[X ],h(ζ k

n ) can be written as a unique polynomial in ζn of degree less than d withinteger coefficients. In particular, we may write mζn (ζ

k) = gk(ζ ) with gk(X)

a polynomial of this form. Since ζ n+kn = ζ k

n , we see that gn+k(X) = gk(X).Now it follows immediately from the Multinomial Theorem and Fermat’s

Little Theorem (a p ≡ a(mod p) for every a ∈ Z, for p a prime) that for anyprime p and for any polynomial f (X) ∈ Z[X ], f (X p) − f (X)p has all ofits coefficients divisible by p. In particular this is true for f (X) = mζn (X),so mζn (X p) − mζn (X)p = pe(X) for some polynomial e(X) ∈ Z[X ]. SettingX = ζ , we see that gp(ζ ) = mζn (ζ

p) = pe(ζ p). Now e(ζ p) can be writtenas a polynomial in ζ of degree less than d with integer coefficients in a uniqueway, and gp(ζ ) is also such a polynomial, so gp(ζ ) = pe(ζ p). Thus we seethat the coefficients of gp(X) are all divisible by p.

There are only finitely many distinct gk(X), so there is an upper bound Afor the absolute values of all of their coefficients. Let p be any prime with p >

A. Then we see that gp(X) is the zero polynomial, so 0 = gp(ζ ) = mζn (ζp

n ),and hence ζn and ζ

pn have the same minimum polynomial mζn (X) (as mζn (X)

is irreducible). We may iterate this argument to conclude that mζ (ζs) = 0 for

any s not divisible by any prime p ≤ A.Let r be any integer relatively prime to n. Let p1, . . . , pt be the primes

less than or equal to A that do not divide r , and set s = r + np1 · · · pt . Thens is such an integer. (If p ≤ A is a prime, then either p is equal to pi forsome i , in which case p divides the second term but not the first, or it is not, inwhich case p divides the first term but not the second.) Then s ≡ r(mod n), so


mζn (ζr ) = mζn (ζ

s) = 0. Thus every primitive nth root of 1 is a root of mζn (X)

and hence n(X) = mζn (X) is irreducible. ��Corollary 4.2.7. (1) mζn (X) = n(X) and (Q(ζn)/Q) = ϕ(n).

(2) Gal(Q(ζn)/Q) ∼= (Z/nZ)∗. In particular, Gal(Q(ζn)/Q) is abelian.

Proof. (1) Since n(ζn) = 0, mζn (X) divides n(X). But n(X) is irreducibleby Theorem 4.2.6 so they are equal. Then (Q(ζn)/Q) = deg mζn (X) = ϕ(n)

where the first equality is Proposition 2.4.6 (2) and the second is Lemma 4.2.3.(2) For any k with (k, n) = 1, ζ k

n is also a primitive nth root of 1, i.e, aroot of mζn (X), so by Lemma 2.6.1 there is a map σk : Q(ζn) → Q(ζn) withσk(ζn) = ζ k

n and σ | Q = id, and σk is determined by its value on ζn . ��Corollary 4.2.8. Let m and n be positive integers and set d = gcd(m, n) and� = lcm(m, n). Then Q(ζm)Q(ζn) = Q(ζ�) and Q(ζm) ∩ Q(ζn) = Q(ζd).

In particular, if m and n are relatively prime, Q(ζm)Q(ζn) = Q(ζmn) andQ(ζm) ∩ Q(ζn) = Q, i.e., Q(ζm) and Q(ζn) are disjoint extensions of Q.

Proof. As ζm = ζ�/m� and ζn = ζ

�/n� , clearly Q(ζ�) ⊇ Q(ζm)Q(ζn). Now

�/m and �/n are relatively prime, so let s and t be integers with s(�/m) +t (�/n) = 1, so 1/� = s/m + t/n. Then ζ� = ζ s

mζ tn so ζ� ∈ Q(ζm)Q(ζn) and

Q(ζ�) = Q(ζm)Q(ζn).Observe that, for any integers r and s with r dividing s, Q(ζr ) ⊆ Q(ζs), so

(Q(ζs)/Q(ζr ))(Q(ζr )/Q) = (Q(ζs)/Q), and hence, by Corollary 4.2.7,

(Q(ζs)/Q(ζr )) = ϕ(s)/ϕ(r).

As ζd = ζm/dm and ζd = ζ

n/dn , clearly Q(ζd) ⊆ Q(ζm) ∩ Q(ζn). Now, by

Corollary 3.4.3 and the above observation,

(Q(ζn)/(Q(ζm) ∩ Q(ζn))) = (Q(ζm)Q(ζn)/Q(ζm))

= (Q(ζ�)/Q(ζm))

= ϕ(�)/ϕ(m).

Furthermore, again using the above observation,

(Q(ζn)/(Q(ζm) ∩ Q(ζn)))((Q(ζm) ∩ Q(ζn))/Q(ζd)) = (Q(ζn)/Q(ζd))

= ϕ(n)/ϕ(d),

so

((Q(ζm) ∩ Q(ζn))/Q(ζd)) = (ϕ(n)/ϕ(d))/(ϕ(�))/ϕ(m))

= (ϕ(m)ϕ(n))/(ϕ(d))ϕ(�)).

4.3 Solvable Extensions and Solvable Groups 97

But it is a fact from elementary number theory that for any two positiveintegers m and n, ϕ(m)ϕ(n) = ϕ(d)ϕ(�), so Q(ζm) ∩ Q(ζn) = Q(ζd), asclaimed. ��

We conclude this section by giving a formula for n(X).

Proposition 4.2.9. For any positive integer n, the n-th cyclotomic polynomialn(X) is given by

n(X) =∏d|n

(Xd − 1)μ(n/d),

where μ is the Mobius μ function, defined by μ( j) = 0 if j is divisible bythe square of a prime, and otherwise μ( j) = (−1)i where i is the number ofdistinct prime factors of j .

Proof. Since, by Lemma 4.2.4 (3), Xn − 1 = ∏d|n d(X), the result fol-

lows immediately from the multiplicative version of the Mobius InversionFormula. ��

4.3 Solvable Extensions and Solvable Groups

In this section, we show that an equation is solvable by radicals if and onlyif its group is a solvable group. Of course, we must begin by defining theseterms. We are most interested in the case F = Q, but the proofs apply moregenerally.

We develop the necessary field theory and the connections between fieldtheory and group theory here. We refer the reader to Section A.1 where wedevelop the necessary group theory.

Definition 4.3.1. (1) A polynomial f (X) ∈ F[X ] is a radical polynomial iff (X) = Xn − a for some positive integer n and element a of F. An extensionE of F is a radical extension (or an extension by radicals) if E = F(α) for α aroot of a radical polynomial.

(2) An equation f (X) = 0, f (X) ∈ F[X ], is solvable by radicals if thereis a sequence of extensions E0 = F ⊆ E1 ⊆ · · · ⊆ Ek with Ei the splittingfield of a radical polynomial fi (X) ∈ Ei−1(X) and with Ek ⊇ E, a splittingfield of f (X). �

(Radical polynomial is not a standard term, but radical extension is.)Before proving our main theorem, Theorem 4.3.4, let us observe a special

case.


Lemma 4.3.2. Let f (X) = Xn −a ∈ F[X ], let E be the splitting field of f (X),and let G = Gal(E/F). Then G is a solvable group.

Proof. E = F(ζn,n√

a). Let B = F(ζn). Then F ⊆ B ⊆ E. Let G1 =Gal(E/B). Now B is a Galois extension of F, so G1 is a normal subgroup of G,and G/G1 is isomorphic to Gal(B/F) by the FTGT. But G/G1 is abelian, byLemma 3.7.3 if F = Q, and by Lemma 3.7.3 and Corollary 3.4.6 in general,and G1 is abelian by Proposition 3.7.6. Then G is solvable, as we see immedi-ately from Definition A.1.1. ��

Let us record a generalization of Lemma 4.3.2 that will help us in the proofof Theorem 4.3.4.

Lemma 4.3.3. Let f (X) = ∏tj=1(Xn j − a j ) with a1, . . . , a j ∈ F, let E be the

splitting field of f (X) over F, and let G = Gal(E/F). Then G is a solvablegroup.

Proof. Define a sequence of fields D0 ⊆ · · · ⊆ Dt inductively, as follows. LetD0 = F and, if D j−1 is defined, let D j be the splitting field of (Xn j − a j ) ∈D j−1[X ]. Note that Dt = E. Let G j = Gal(Dt− j/D0) for j = 0, . . . , t . ThenG = G0 ⊇ · · · ⊇ Gt = {1}. Also, G j−1/G j = Gal(Dt− j+1/Dt− j ). But thisgroup is solvable by Lemma 4.3.2, so, by Lemma A.1.3, G is solvable. ��

Now for our main theorem.

Theorem 4.3.4. Let f (X) ∈ F[X ] and let E be a splitting field of f (X). Thenf (X) = 0 is solvable by radicals if and only if G = Gal(E/F) is a solvablegroup.

Proof. First, assume G = Gal(E/F) is solvable. Let deg f (X) = m and setn = m!.

Case I: F ⊇ Q(ζn). Let G = G0 ⊇ · · · ⊇ Gk = {1} with Gi a normalsubgroup of Gi−1 and Gi−1/Gi cyclic (see Lemma A.1.3), say of order mi .Note mi ≤ m so in particular mi divides n. Let Bi = Fix(Gi ), and note thatB0 = F and Bk = E. Then Bi is a cyclic extension of Bi−1 of degree mi , andBi−1 ⊇ Q(ζn) ⊇ Q(ζmi ), so Bi is the splitting field of Xmi − ai , for someai ∈ Bi−1, by Proposition 3.7.7.

Case II: The general case. The equation Xn − 1 is obviously solvable byradicals. Let B = F(ζn), the splitting field of this polynomial. Let EB be thesplitting field of f (X) ∈ B[X ]. By Corollary 3.4.6, Gal(EB/B) is isomorphicto a subgroup of Gal(E/F). But, by Lemma A.1.4 (1), a subgroup of a solvablegroup is solvable, so we are reduced to case I.

4.3 Solvable Extensions and Solvable Groups 99

Conversely, assume that f (X) = 0 is solvable by radicals. Then there isa sequence of fields F = B0 ⊆ B1 ⊆ · · · ⊆ Bk with Bi the splitting field ofXmi −ai , ai ∈ Bi−1, and with Bk ⊇ E. Then Gal(Bi/Bi−1) is a solvable group,by Lemma 4.3.2.

Suppose for the moment that k = 2. Then we have that Gal(B2/B1) andGal(B1/B0) are solvable groups, and that B2/B1 and B1/B0 are Galois exten-sions. If we knew that B2/B0 was a Galois extension, we could apply LemmaA.1.4 (3) to conclude that G = Gal(B2/B0) was a solvable group. However,a sequence of Galois extensions may not be Galois (and indeed, one can con-struct such an example precisely in this case). Thus in order to proceed wemust enlarge the fields we are dealing with.

We proceed inductively. We begin by setting B′0 = B0 = F and B′

1 =B1, and observe that B′

1 is a Galois extension of F. Suppose B′i−1 is de-

fined and is a Galois extension of F. We let B′i be the splitting field of∏

σ∈Gal(B′i−1/B′

0)(Xmi − σ(ai )) ∈ B′

i−1[X ]. Observe that this polynomial is in-variant under Gal(B′

i−1/B′0) so it is in fact in B′

0[X ] = F[X ]. Thus B′i is the

splitting field of a separable polynomial in F[X ] so it is a Galois extension ofF. Thus we obtain a sequence of fields F = B′

0 ⊆ B′1 ⊆ · · · ⊆ B′

k with eachB′

i a Galois extension of F and with B′k ⊇ Bk ⊇ E.

Let G = Gal(B′k/F) and let Gi = Gal(B′

k−i/B′0) for i = 0, . . . , k. Then

G = G0 ⊇ · · · ⊃ Gk = {1}. Also, Gi−1/Gi = Gal(B′k−i+1/B′

k−i ). By Lemma4.3.3, Gi−1/Gi is solvable for each i = 1, . . . , k. Then, by Lemma A.1.3,G = Gal(B′

k/F) is solvable. Now F ⊆ E ⊆ B′k and E is a Galois extension

of F, so Gal(E/F) = Gal(B′k/F)/ Gal(B′

k/E) is a quotient of a solvable groupand hence is solvable, by Lemma A.1.4 (2). ��Corollary 4.3.5 (Abel). The general equation of degree n is not solvable byradicals for n ≥ 5.

Proof. Let fn(X) ∈ F[X ] be a polynomial with splitting field E, whereGal(E/F) = Sn , the symmetric group on n elements. Such a polynomial existsby Lemma 3.1.2, and in fact we have explicitly exhibited one in Remark 3.1.5.Now, by Corollary A.3.6, Sn is not solvable for n ≥ 5. Thus, for n ≥ 5, fn(X)

is not solvable by radicals, by Theorem 4.3.4. ��While we credit Theorem 4.3.4 to Galois, in fact he did not state it that

way. We shall now derive Galois’s original result. One the one hand, this is ofhistorical interest, but on the other hand, it provides an easy way of exhibitinga polynomial in Q[X ] whose Galois group is not solvable.

Theorem 4.3.6 (Galois). Let f (X) ∈ F[X ] be an irreducible polynomial ofprime degree p. Then f (X) = 0 is solvable by radicals if and only if all of itsroots are rational functions of any two of them.


Proof. By Theorem 4.3.4 we know that f (X) = 0 is solvable by radicals ifand only if G = Gal(E/F) is a solvable group, where E is a splitting field off (X). Since f (X) is irreducible, G acts transitively on the roots of f (X).

First, suppose that all of the roots of f (X) are rational functions of anytwo of them. Regard G as a subgroup of the symmetric group Sp by its actionpermuting the roots of f (X), and let σ ∈ G. If α1 �= α2 are roots, then for anyε ∈ E, σ(ε) is determined by σ(α1) and σ(α2). There are at most p(p − 1)

choices for σ(α1) and σ(α2), so G has order at most p(p − 1), and so byCorollary A.1.9 G is solvable.

Conversely, suppose that G is solvable. Let α1 and α2 be any two distinctroots of f (X), and let B = F(α1, α2) ⊆ E. We need to show that B = E.

We apply the Fundamental Theorem of Galois Theory (FTGT), Theorem2.8.8. B is a field intermediate between F and E, so G ′ = Gal(E/B) ⊆ G =Gal(E/F). By definition, Gal(E/B) = {σ ∈ G | σ(β) = β for every β ∈ B}.But the FTGT establishes a 1−1 correspondence between subgroups of G andfields intermediate between F and E, so B = E if and only if G ′ = {id}. Weproceed to show this.

Let σ ∈ G. Note that σ ∈ G ′ if and only if σ(α1) = α1 and σ(α2) = α2.By Proposition A.1.7 (and using the notation there), we may identify G

with G H for some subgroup H of F∗p, and α1 and α2 with

[i11

]and

[i21

]for

some i1 �= i2.Let σ = [

h n0 1

] ∈ G H with σ(α1) = α1 and σ(α2) = α2. Then[

i1

1

]=

[h n0 1

] [i1

1

]=

[hi1 + n

1

],

[i2

1

]=

[h n0 1

] [i2

1

]=

[hi2 + n

1

],

and these two equations readily imply h = 1, n = 0, i.e., σ = id. Thus we seethat G ′ = {id}, completing the proof. ��Corollary 4.3.7. Let F ⊆ R and let f (X) ∈ F[X ] be an irreducible polyno-mial of prime degree p with k real roots, 1 < k < p. Then f (X) = 0 is notsolvable by radicals.

Proof. Suppose f (X) has real roots α1 �= α2 and a nonreal root β. Then β isnot a rational function of α1 and α2, so, by Theorem 4.3.6, f (X) = 0 is notsolvable by radicals. ��Remark 4.3.8. In Example 4.7.5 we will see that, for each odd prime p, thereis an irreducible polynomial f p(X) ∈ Q[X ] with p − 2 real roots. Thus, byCorollary 4.3.7, f p(X) is not solvable by radicals for p ≥ 5. But in fact wewill show the stronger result that the Galois group of f p(X) over Q is iso-morphic to Sp, thus providing an example of Corollary 4.3.5 over Q. We will

4.4 Geometric Constructions 101

further provide an example of such a polynomial fn(X) over Q for every n ≥ 5in Proposition 4.7.10, but the construction there is considerably more compli-cated. �

4.4 Geometric Constructions

In this section we will investigate the question of constructibility by straight-edge and compass. We will see that the three classical questions of Greekantiquity (trisecting the angle, duplicating the cube, and squaring the circle)are unsolvable by these methods. We will also see when it is possible to con-struct a regular n-gon by these methods, a result originally due to Gauss. Inparticular, we will show how to construct a regular 17-gon.

We begin by drawing an arbitrary line L , and choose two distinct points Oand A on it. We normalize coordinates by letting O be the origin and A be thepoint (1, 0), so in particular the line segment O A has length 1. Thus the lineL is the x-axis.

We recall that, using straightedge and compass, it is possible to construct aline through a given point parallel to, or perpendicular to, a given line. Thus, inparticular, we may construct the y-axis and hence we have the usual Cartesiancoordinates in the plane.

There are two ways to regard points in the plane. We may consider that apoint P has coordinates (x, y), or that P is represented by the complex numberz = x + iy. If we have constructed the point P represented by z = x + iy, thenwe may drop perpendiculars to obtain the real and imaginary parts x and y ofP; conversely, given real numbers x and y we may certainly obtain the point(x, y) corresponding to z = x + iy. Also, all arithmetic operations on complexnumbers (addition, subtraction, multiplication, division) may be performed bydoing arithmetic operations on their real and imaginary parts. Thus we shallpass freely between these two different descriptions.

Our first main result is the following:

Theorem 4.4.1. The complex number z can be constructed by straightedgeand compass if and only if z is algebraic over Q and there is a sequence offields

F0 = Q ⊂ F1 ⊂ · · · ⊂ Fk

with Q(z) ⊆ Fk and with (Fi/Fi−1) = 2 for each i = 1, . . . , k.

Proof. First, we observe that we may perform all arithmetic operations withstraightedge and compass. Clearly we can add and subtract (positive) real num-bers x and y. We may multiply them as follows: Let the line segment AB have


length x and extend AB to a line L . Let M be another line through A and letC and D be points on the same side of A with the line segments AC and ADhaving lengths 1 and y, respectively. Construct the line N through D paral-lel to the line segment C B and let this line intersect the line L in the pointE . Then, by similar triangles, the line segment AE has length xy. To dividewe perform a similar construction, this time letting N be the line through Cparallel to the line segment DB, and then AE has length x/y.

We also observe that we may take the square root of a (positive) real num-ber x using straightedge and compass. To do so, let line segment AB havelength x and extend AB to a line L . Let C be a point on the opposite side of Afrom B with the length of the line segment AC equal to 1. Construct a semi-circle with diameter C B, and a line M perpendicular to C B at A. Let D be thepoint at which the line M intersects this semicircle. Then, by similar triangles,AD has length

√x .

With this in hand, we turn to the proof of the theorem.First, suppose there is a sequence of fields as indicated. By our first obser-

vation, every x ∈ Q = F0 can be constructed. Since (F/F0) = 2, F1 = F0(α1)

where α1 is a root of an irreducible quadratic equation with coefficients in F0.Then, by the quadratic formula, F0(α1) = F0(

√a1) for some a1 ∈ F0, and by

our second observation,√

a1 can be constructed, and hence every x ∈ F1 canbe constructed. Then proceed by induction.

Conversely, suppose a point can be constructed with a straightedge andcompass. Then it is obtained by a succession of the following operations:

(1) Finding the intersection of two lines.(2) Finding the intersection of a circle and a line.(3) Finding the intersection of two circles.In case of operation (1), let the two lines be given by the equations

a1x + b1 y = c1 and a2x + b2 y = c2 with a1, a2, b1, b2, c1, c2 ∈ F for somefield F. Then the solution (x, y) has x ∈ F and y ∈ F. In case of operation(2), let the circle be given by (x − a1)

2 + (y − b1)2 = c2

1 and the line begiven by a2x + b2 y = c2 with a1, a2, b1, b2, c1, c2 ∈ F. Solving for y in termsof x (or vice versa) in the linear equation, substituting in the quadratic equa-tion, and solving, shows x ∈ F(d) and y ∈ F(d) where (F(d)/F) = 1 or 2.In case of operation (3), let the two circles have equations (x − a1)

2 +(y − b1)

2 = c21 and (x − a2)

2 + (y − b2)2 = c2

2 with a1, a2, b1, b2, c1, c2 ∈ F.Subtracting the second equation from the first gives a linear equation a3x +b3 y = c3 for some a3, b3, c3 ∈ F, so combining the equation of the first circle(say) with this linear equation, we are reduced to operation (2). This yields thetheorem. ��


Corollary 4.4.2. It is impossible to perform the following constructions usingstraightedge and compass:

(1) (Trisecting the angle) Trisecting an arbitrary angle.(2) (Duplicating the cube) Constructing a cube with volume twice that of

a given cube.(3) (Squaring the circle) Constructing a square with area that of a given

circle.

Proof. (1) Given an angle θ , let one side of the angle be the x-axis and let theother side of the angle intersect the unit circle at P . Then P has coordinates(cos θ, sin θ). Thus we may construct the angle θ/3 if and only if we mayconstruct the real number cos θ/3. Recall the formula cos 3ϕ = 4 cos4 ϕ −3 cos ϕ. Letting ϕ = θ/3 and x = cos ϕ, we see that x satisfies the cubicequation 4x3 − 3x = cos θ . We may certainly construct an angle θ = π/3 (=60◦), and cos(π/3) = 1

2 . Thus x is a root of the polynomial f (X) = 8X3 −6X − 1. Using Lemma 4.1.2, it is easy to check that f (X) has no linear factorin Q(X), and hence is irreducible, so (Q(x)/Q) = 3. But then we cannot haveQ(x) ⊆ Fk with Fk as in Theorem 4.4.1, as (Fk/Q) is a power of 2 and hencenot divisible by 3.

(2) Trivially, we may construct a segment of length the side of a cube ofvolume 1. But then to construct a segment of length the side of a cube ofvolume 2 would be to construct x = 3

√2. But (Q(

3√

2)/Q) = 3 does not divideany power of 2, so this is impossible as in part (1).

(3) Trivially, we may construct a segment of length the radius of a circle ofarea π . But then to construct a segment of length the side of a square of area π

would be to construct x = √π . But it is a famous theorem of Lindemann that

π is transcendental (i.e., not algebraic, or equivalently that π and hence√

π

does not satisfy any algebraic equation) so (Q(√

π)/Q) is infinite and hence√π is not an element of any finite extension of Q (regardless of degree). ��

We have an equivalent formulation of the condition in Theorem 4.4.1.

Lemma 4.4.3. For a complex number z, the following are equivalent:(1) There is a sequence of fields F0 = Q ⊂ F1 ⊂ · · · ⊂ Fk with Q(z) ⊆ Fk

and with (Fi/Fi−1) = 2 for each i = 1, . . . , k.(2) If E is a splitting field of mz(X), then (E/Q) is a power of 2.

Proof. (2) implies (1): Let G = Gal(E/Q) and let G have order 2k . By Corol-lary A.2.3 there is a sequence of subgroups G = G0 ⊃ G1 ⊃ · · · ⊃ Gk = {1}with [G : Gi ] = 2i . Let Fi = Fix(Gi ). Then {Fi } is as required.

(1) implies (2): We prove this by induction on k. For k = 0 it is triv-ial and for k = 1 it is immediate, as every quadratic extension is a split-ting field. Let (Fk/Fk−1) = 2. Then Fk = Fk−1(αk) for some αk ∈ Fk with


deg mαk (X) = 2. In the course of the proof we shall have occasion to replacethe field Fk−1 by another field Ek−1. Observe that for any field Ek−1 ⊇ Fk−1,(Ek−1(αk)/Ek−1) = 1 or 2 by Lemma 2.4.7.

Let E0 = F0 and E1 = F1. If αk ∈ Ek−1, then Ek = Ek−1(αk) = Ek−1 andthere is nothing to do. Otherwise, let G = Gal(Ek−1/E0) and let

f (X) =∏σ∈G

σ(mαk (X)).

Then f (X) is invariant under G, so f (X) ∈ E0[X ]. Let Ek be a splittingfield of f (X) containing Fk . Then f (X) has roots β1 = αk, . . . , βm . Nowmαk (X) has degree 2, so σ(mα(X)) has degree 2 for each σ ∈ G, and hence(Ek−1(βi )/Ek−1) = 2 for each i . Now consider

Ek−1 ⊆ Ek−1(β1) ⊆ Ek−1(β1, β2) ⊆ · · · ⊆ Ek−1(β1, . . . , βm) = Ek .

Set E′k−1 = Ek−1(β1, . . . , βi−1). By Remark 2.3.6, Ek−1(β1, . . . , βi ) =

Ek−1(β1, . . . , βi−1)(βi ) = E′k−1(βi ) is the composite E′

k−1Ek−1(βi ), so, byLemma 2.3.8,

(Ek−1(β1, . . . , βi )/Ek−1(β1, . . . , βi−1)) = (E′k−1(βi )/E′

k−1)

≤ (Ek−1(βi )/Ek−1) = 2,

so each successive extension is of degree 1 (i.e., is trivial) or of degree 2.Thus (Ek/Q) is of degree a power of 2, and is a splitting field, hence a Galoisextension of Q. Now z ∈ Ek , so mz(X) splits in Ek and hence mz(X) has asplitting field E with Q ⊆ E ⊆ Ek . Then (E/Q) divides (Ek/Q), so (E/Q) isa power of 2. ��Remark 4.4.4. For an arbitrary prime p, the implication (2) implies (1) ofLemma 4.4.3 holds, with the identical proof, but the implication (1) implies(2) does not. The proof breaks down as Lemma 2.3.8 merely guarantees adegree between 1 and p, but it may be strictly between 1 and p. A counter-example to the implication (1) implies (2) of Lemma 4.4.3 is given by z = 3

√2.

Then (Q(z)/Q) = 3 but mz(X) = X3 − 2, whose splitting field E over Q has(E/Q) = 6, which is not a power of 3. (Compare Example 2.3.7.) �

We now do a bit of elementary number theory. Let n = 2t + 1 for somepositive integer t . If t has an odd factor r > 1, then, writing t = rs, n isdivisible by 2s + 1 (as we see by substituting X = 2s in the algebraic identityXr +1 = (X +1)(Xr−1 − Xr−2 +· · ·+1), valid for any odd r ). Thus the onlypossible primes of this form are when t = 2k for some k. Set Fk = 22k + 1.


If Fk is prime, it is called a Fermat prime. This terminology is due to thefact that Fermat claimed Fk is prime for every k. Now F0 = 3, F1 = 5,F2 = 17, F3 = 257, and F4 = 65537 are indeed prime, but Euler discoveredthat F5 = 4294967297 is divisible by 641 and hence is composite. In fact,there is no known value of k > 4 for which Fk is prime.

Here is our second main result:

Theorem 4.4.5 (Gauss). A regular n-gon is constructible by straightedge andcompass if and only if n is of the form

n = 2k p1 · · · p j

with {pi } distinct Fermat primes.

Proof. Clearly a regular n-gon is constructible if and only if a regular n-goninscribed in a circle of radius 1 is constructible, and this is true if and only ifits vertices are constructible, and this is true if and only if ζn = exp(2π i/n) isconstructible.

By Theorem 4.4.1 and Lemma 4.4.3, this is true if and only if the split-ting field of mζn (X) has degree a power of 2. But this splitting field is simplyQ(exp(2π i(n)), the nth cyclotomic field, which has degree ϕ(n) by Corollary4.2.7.

If n = 2a pb1 pc

2 · · · , then ϕ(n) = 2a−1(p1 − 1)pb−11 (p2 − 1)pc−1

2 · · · , so wesee that ϕ(n) can be a power of 2 if and only if b = c = · · · = 1 and each ofp1, p2, . . . is more than a power of 2, and hence a Fermat prime. ��Example 4.4.6. (1) Everyone knows how to construct an equilateral triangle.But in any case, ζ3 satisfies the equation X2 + X + 1 = 0. In fact, construct-ing ζ3 = (−1 + i

√3)/2 gives a construction of an equilateral triangle that

is different from the usual one. It is the following (using the notation at thebeginning of this section): Extend the line segment AO so that it intersects theunit circle at the point D. Construct the midpoint E of the line segment O D,and then construct the line M perpendicular to the segment O D at E . Let thisline intersect the unit circle at the points B and C . Then ABC is an equilateraltriangle.

(2) We essentially showed how to construct a regular pentagon in Example1.1.4. To review that, let θ = ζ5 + ζ 4

5 = ζ5 + ζ−15 . Then θ satisfies the equation

X2 − X − 1 = 0 and ζ5 satisfies the equation X2 − θ X + 1 = 0.(3) We show how to construct a regular 17-gon. Let ζ = ζ17 =

exp(2π i/17). Let G = G0 = Gal(Q(ζ )/Q). Then we know that G is cyclicof order 16. Write G = G0 ⊃ G1 ⊃ G2 ⊃ G3 ⊃ G4 = {1} where[G : Gi ] = 2i , and note that Gi is cyclic of order 24−i . Let Fi = Fix(Gi )


so F0 = Q and F4 = Q(ζ ). G is isomorphic to F∗17, and direct calcula-

tion shows that 3 is a generator of this group. (In general, a generator ofF∗

p is called a primitive root mod p.) Let σi (ζ ) = ζ 3i, i = 0, . . . , 15. Then

G0 = {σ0, σ1, . . . , σ15}, G1 = {σ0, σ2, . . . , σ14}, G2 = {σ0, σ4, σ8, σ12},G3 = {σ0, σ8}, and G4 = {σ0}. Also, (Fi/Fi−1) = 2 for i = 1, . . . , 4. WriteE = F4 = Q(ζ ).

In order to simplify our computations, we recall the following elementaryfact: If r and s are the roots of the quadratic equation X2 − aX + b = 0, thenr + s = −a and rs = b. (Proof: (X − r)(X − s) = X2 − (r + s)X + rs.)

We work from the bottom up. (F1/F0) = 2 so F1 = F0(α) for any α ∈ Efixed under G1 but not under G0, i.e., any α with σ2(α) = α but σ1(α) �= α.Let

α0 = ∑7i=0 ζ 32i = ζ + ζ 9 + ζ 13 + ζ 15 + ζ 16 + ζ 8 + ζ 4 + ζ 2,

α1 = ∑7i=0 ζ 32i+1 = ζ 3 + ζ 10 + ζ 5 + ζ 11 + ζ 14 + ζ 7 + ζ 12 + ζ 6.

If α = α0 or α1, then σ2(α) = α but σ1(α) �= α. Direct calculation showsα0 + α1 = −1 and α0α1 = −4, so α0 and α1 are the roots of X2 + X − 4 = 0.

Now (F2/F1) = 2 so F2 = F1(β) for any β ∈ E fixed under G2 but notG1, i.e., any β with σ4(β) = β but σ2(β) �= β. Let

β0 = ζ + ζ 13 + ζ 16 + ζ 4,β1 = ζ 3 + ζ 5 + ζ 14 + ζ 12,β2 = ζ 9 + ζ 15 + ζ 8 + ζ 2,β3 = ζ 10 + ζ 11 + ζ 7 + ζ 6.

If β = β0, β1, β2, or β3, then σ4(β) = β but σ2(β) �= β. Then β0 +β2 =α0, β0β2 = −1, and β1β3 = α1, β1β3 = −1, so β0 and β2 are the roots ofX2 − α0 X − 1 = 0 and β1 and β3 are the roots of X2 − α1 X − 1 = 0.

Next (F3/F2) = 2 and F3 = F2(γ ). Let

γ0 = ζ + ζ 16,γ4 = ζ 13 + ζ 4.

If γ = γ0 or γ4, then σ8(γ ) = γ but σ4(γ ) �= γ . Then γ0 + γ4 = β0,γ0γ4 = β1, so γ0 and γ4 are the roots of X2 − β0 X + β1 = 0.

Finally, let δ0 = ζ and δ8 = ζ 16. If δ = δ0 or δ8, then σ8(δ) �= δ. SetF4 = F3(δ0) = F3(ζ ). Note that ζ satisfies ζ + ζ 16 = γ0, ζ ζ 16 = 1, so ζ is aroot of X2 − γ0 X + 1 = 0.

Summarizing, we construct ζ = ζ17 as follows:

(1) Solve the quadratic X2 + X − 4 = 0; call the roots α0 and α1.

4.5 Quadratic Extensions of Q 107

(2a) Solve the quadratic X2 − α0 X − 1 = 0; call the roots β0 and β2.(2b) Solve the quadratic X2 − α1 X − 1 = 0; call the roots β1 and β3.

(3) Solve the quadratic X2 − β0 X + β1 = 0; call the roots γ0 and γ4.(4) Solve the quadratic X2 −γ0 X +γ4 = 0; its roots are ζ = δ0 and ζ 16 = δ8.

With enough patience (and careful attention to sign) the reader may ob-tain an explicit formula for ζ which involves square roots nested four deep.(To start off, α0 = (−1 + √

17)/2 and α1 = (−1 − √17)/2, showing that

Q(√

17) ⊂ Q(ζ17). Compare Corollary 4.5.2.) �

4.5 Quadratic Extensions of Q

We recall a bit of elementary number theory. Let p be a prime. Identifying Fp

with Z/pZ, we know that G = F∗p = {1, . . . , p − 1} is a cyclic group, and

a generator r of this group is called a primitive root mod p. For example, 3 isa primitive root mod 7 as 31 ≡ 3 (mod 7), 32 ≡ 2 (mod 7), 33 ≡ 6 (mod 7),34 ≡ 4 (mod 7), 35 ≡ 5 (mod 7), and 36 ≡ 1 (mod 7). On the other hand,2 is not a primitive root mod 7 as 23 ≡ 1 (mod 7). (We must be careful notto confuse a primitive root mod p with a primitive pth root of 1. Observe thatζ k

p is a primitive pth root of 1 for any k not divisible by p.) Observe also that(Z/4Z)∗ = {1, 3}, so 3 is a primitive root mod 4.

If p is an odd prime, then G is cyclic of even order p − 1, so it has aunique subgroup H of index 2. Let π : G → G/H ∼= Z/2Z = {0, 1} bethe quotient map. For an integer k relatively prime to p, we regard k as anelement of Z/pZ. Then χp, the quadratic residue character mod p, is definedby χp(k) = (−1)π(k). (For fixed p, we shall write χ for χp.)

There is a more traditional definition of the quadratic residue character.Recalling that G is a multiplicative group, the subgroup H consists of the(quadratic) residues, i.e., of those elements of G that are squares, and its com-plement consists of the (quadratic) nonresidues, i.e., of those elements of Gthat are nonsquares. Then χ(k) = 1 if k is a residue and χ(k) = −1 if k is anonresidue.

Furthermore, if r is a generator of G, i.e., a primitive root mod p, thenχ(k) = 1 if k is an even power of r and χ(k) = −1 if k is an odd power of r(and this is independent of the choice of r ).

For example, if p = 7, 1 = (±1)2, 4 = (±2)2, and 2 = (±3)2 are theresidues and 3, 5, and 6 are the nonresidues. Choosing the primitive root 3, wesee that 1 = 30, 2 = 32, and 4 = 34, while 3 = 31, 5 = 35, and 6 = 33.

It is then easy to see that the quadratic residue character has the followingproperties:


(1) χ( jk) = χ( j)χ(k). (χ is a homomorphism to the multiplicativegroup {±1}.)

(2)∑p−1

k=1 χ(k) = 0. (There are (p − 1)/2 residues and (p − 1)/2 non-residues.)

(3) χ(−1) = (−1)(p−1)/2 = 1 if p ≡ 1 (mod 4) and = −1 if p ≡3 (mod 4). (If r is a primitive root mod p, let a = r (p−1)/2. Then a �= 1 buta2 = 1. On the other hand, the equation X2 − 1 = 0 has only two solutions inFp, X = ±1. Hence a = −1. Thus if p ≡ 1 (mod 4), (p − 1)/2 is even, a isan even power of r , and χ(a) = 1, while if p ≡ 3 (mod 4), (p − 1)/2 is odd,a is an odd power of r , and χ(a) = −1.)

Theorem 4.5.1. Let E be a quadratic extension of Q. Then E is intermediatebetween Q and some cyclotomic field. In particular:

(1) Q(√−1) = Q(ζ4).

(2) Q(√

2) ⊂ Q(ζ8) and Q(√−2) ⊂ Q(ζ8).

(3) If p is a prime congruent to 1 modulo 4, then Q(√

p) ⊆ Q(ζp).(4) If p is a prime congruent to 3 modulo 4, then Q(

√−p) ⊆ Q(ζp).

Proof. Observe that if Q(√

a) ⊆ Q(ζm) and Q(√

b) ⊆ Q(ζn), then Q(√

ab) ⊆Q(

√a,

√b) ⊆ Q(ζm, ζn) ⊆ Q(ζmn), so in order to prove the general statement

it suffices to prove the particular statements (1), (2), (3), and (4).(1) is obvious.(2) is direct calculation: ζ8 = (1 + i)/

√2.

We prove (3) and (4) simultaneously. Actually, we present two proofs.Let p be an odd prime and let

Sp =p−1∑k=1

χ(k)ζ kp .

Then

S2p =

p−1∑k=1

p−1∑j=1

χ(k)ζ kpχ( j)ζ j

p =p−1∑k=1

p−1∑j=1

χ(k j)ζ k+ jp .

Set j ≡ km (mod p) and notice that as j runs over the nonzero congruenceclasses mod p, so does m. Also, χ(k j) = χ(k2m) = χ(m). Thus

S2p =

p−1∑k=1

p−1∑m=1

χ(m)ζ k+kmp

=p−1∑m=1

χ(m)

p−1∑k=1

(ζ 1+mp )k .


Now 1 + ζp + · · · + ζp−1p = 0, so as long as m �= p − 1, the inner sum is

−1. If m = p − 1, then the inner sum is of course p − 1.Thus

S2p =

(−

p−2∑m=1

χ(m)

)+ (p − 1)χ(p − 1).

But∑p−1

m=1 χ(m) = 0, so the first term is +χ(−1) and we see

S2p = pχ(−1),

as required.

Our second proof uses more theory and a minimum of calculation. LetG = Gal(Q(ζp)/Q) and recall that G is generated by σ0, where σ0(ζp) = ζ r

pwith r a primitive root mod p. Since Q(ζp) is a Galois extension of Q, anyα ∈ Q(ζp) with σ0(α) = α (and hence with σ(α) = α for every σ ∈ G) is infact an element of Q.

Now (Q(ζp)/Q) = p − 1 and Q(ζp) is certainly spanned by {1, ζp, . . . ,

ζp−1p }. We know one relation among these p elements: 1 + ζp + · · · + ζ

p−1p =

0. Hence {ζp, . . . , ζp−1p } form a basis for Q(ζp), so in particular this set is

Q-linearly independent. Thus if α ∈ Q(ζp) with σ0(α) = α, then, writingα = ∑p−1

i=1 aiζip with ai ∈ Q, we must have a1 = · · · = ap−1 = a, for some a.

(Of course, in this case α = −a ∈ Q.)With these observations in mind, let us set

α0 = α0(p) =(p−1)/2∑

i=1

ζ r2i−2

p ,

α1 = α1(p) =(p−1)/2∑

i=1

ζ r2i−1

p .

Observe that the exponents in α0 are all quadratic residues mod p, and thatthe exponents in α1 are all quadratic nonresidues mod p. Then

α0 + α1 = ζp + · · · ζ p−1p = −1 and σ(α0) = α1, σ (α1) = α0.

Let b = α0α1. Then

σ0(b) = σ0(α0α1) = σ0(α0)σ0(α1) = α1α0 = b,

so b ∈ Q. We now compute b.


First, suppose p ≡ 1 (mod 4). There are (p−1)/2 terms in α0 and (p−1)/2terms in α1, so there are (p − 1)2/4 terms in the product. Each term is ofthe form ζ

jp ζ k

p with j a quadratic residue and k a quadratic nonresidue. Sincep ≡ 1 (mod 4), χ(− j) = χ(−1)χ( j) = χ( j), so j and − j are always eitherboth residues or both nonresidues, thus ζ k

p �= (ζjp )−1 and hence no term ζ

jp ζ k

p

is equal to 1. Then, writing b = ∑p−1i=1 aiζ

ip, we see that

∑ai = (p − 1)2/4

and that a1 = · · · = ap−1 = a, for some a, so∑

ai = (p − 1)a = (p − 1)2/4and hence a = (p − 1)/4. Then

b = ap−1∑i=1

ζ i = ((p − 1)/4)

p−1∑i=1

ζ i = ((p − 1)/4)(−1) = −(p − 1)/4.

Now suppose p ≡ 3 (mod 4). Then χ(− j) = −χ( j), so one of j and − jis always a quadratic residue and the other is a quadratic nonresidue. Henceeach term in α0 “pairs up” with one of the terms in α1 to give a product of1. This takes care of (p − 1)/2 of the (p − 1)2/4 terms. Thus we may writeb = (p − 1)/2 + b′ where b′ is the sum of the remaining (p − 1)2/4 −(p − 1)/2 = (p − 1)(p − 3)/4 terms. We now apply the same logic to b′.Writing b′ = ∑p−1

i=1 a′iζ

ip, we see that

∑a′

i = (p − 1)(p − 3)/4 and thata′

1 = · · · = a′p−1 = a′, for some a′, so

∑a′

i = (p − 1)a′ = (p − 1)(p − 3)/4,and hence a′ = (p − 3)/4. Then

b′ = a′p−1∑i=1

ζ i = ((p − 3)/4)

p−1∑i=1

ζ i = ((p − 3)/4)(−1) = −(p − 3)/4,

and then

b = (p − 1)/2 + b′ = (p − 1)/2 − (p − 3)/4 = (p + 1)/4.

Now

(X − α0)(X − α1) = X2 − (α0 + α1)X + α0α1

= X2 + X + b,

so

(X − α0)(X − α1) = X2 + X − (p − 1)/4 if p ≡ 1 (mod 4),

(X − α0)(X − α1) = X2 + X + (p + 1)/4 if p ≡ 3 (mod 4).


In other words, α0 and α1 are the roots of this quadratic equation. But wemay find these roots by the quadratic formula, and we obtain

{α0, α1} = {(−1 ± √p)/2} if p ≡ 1 (mod 4),

{α0, α1} = {(−1 ± √−p)/2} if p ≡ 3 (mod 4). ��Corollary 4.5.2. For p an odd prime, Q(

√χp(−1)p) is the unique quadratic

field intermediate between Q and Q(ζp).

Proof. Gal(Q/ζp)/Q) ∼= Z/(p − 1)Z contains a unique subgroup of indextwo, so there is a unique quadratic field intermediate between Q and Q(ζp),and we have just identified that field. ��Remark 4.5.3. Let us cite some results from number theory that will enable usto extend Corollary 4.5.2. They are:

(1) If m1 and m2 are relatively prime, then (Z/(m1m2)Z)∗ ∼= (Z/m1Z)∗ ⊕(Z/m2Z)∗. (This follows easily from the Chinese Remainder Theorem.)

(2) For p an odd prime and k a positive integer, (Z/pkZ)∗, a group of orderϕ(pk) = (p − 1)pk−1, is cyclic. (This is usually phrased as saying there is aprimitive root mod pk .)

(3) (Z/4Z)∗ ∼= Z/2Z and for k > 2, (Z/2kZ)∗, a group of order ϕ(2k) =2k−1, is isomorphic to the direct sum of a cyclic group of order 2 and a cyclicgroup of order 2k−2. �Corollary 4.5.4. Let n > 2 be an integer. Define a set A = {ai }i=1,...,t asfollows: If p is an odd prime factor of n, then χp(−1)p ∈ A. If n is divisibleby 4, then −1 ∈ A. If n is divisible by 8, then 2 ∈ A. Then Q(ζn) contains2t − 1 quadratic extensions of Q, and they are Q(

√m) for m any nontrivial

product of distinct elements of A.

Proof. By the Fundamental Theorem of Galois Theory, the quadratic exten-sions of Q contained in Q(ζn) are in 1 − 1 correspondence with the subgroupsof index 2 of Gal(Q(ζn)/Q ∼= (Z/nZ)∗. Now if n = 2e2 3e3 5e5 . . . , (Z/nZ)∗ ∼=(Z/2e2Z)∗ × (Z/3e3Z)∗ × (Z/5e5Z)∗ × · · · . For p odd and k ≥ 1, (Z/pkZ)∗is a cyclic group of even order. Also, (Z/2)∗ is trivial, (Z/4)∗ ∼= Z/2Z, and(Z/2kZ)∗ ∼= (Z/2Z) ⊕ (Z/2k−2Z). Thus (Z/nZ)∗ contains 2t − 1 subgroupsof index 2, where t is as in the statement of the corollary. (A subgroup of in-dex 2 is the kernel of an epimorphism ψ : Gal(Q(ζn)/Q) → Z/2Z and sinceGal(Q(ζn)/Q) is isomorphic to the direct sum of t cyclic groups of even order,there are 2t homomorphisms from Gal(Q(ζn)/Q) to Z/2Z, one of which isthe trivial one.) Since Q(

√m) ⊆ Q(ζn) for each of the 2t − 1 values of m in

the statement of the corollary, by Theorem 4.5.3, we see that these are all thequadratic subfields of Q(ζn). ��


Corollary 4.5.5. Let m be a square-free integer. If m ≡ 1 (mod 4), set m ′ =|m|. Otherwise set m ′ = 4|m|. Then Q(

√m) ⊆ Q(ζn) if and only if n is a

multiple of m ′.

Proof. Corollary 4.5.4 gives all the quadratic subfields of Q(ζn). ��Remark 4.5.6. We have shown in the proof of Theorem 4.5.1 that for an oddprime p, Sp = ±√

χp(−1)p. It is a famous theorem of Gauss that in everycase the sign is positive. It is easy to check that Sp = 2α0(p) + 1, so we seethat α0(p) = (−1 + √

χp(−1)p)/2 in every case as well. �

4.6 Radical Polynomials and Related Topics

In this section we investigate radical polynomials f (X) ∈ Q[X ], i.e., poly-nomials of the form Xn − a, their associated radical extensions Q( n

√a), and

their splitting fields Q( n√

a, ζn). The results we obtain are interesting in theirown right, and, as we shall see, also have applications to cyclotomic fields andKummer fields.

First we determine the Galois groups of their splitting fields. We remindthe reader of our discussion of primitive roots at the beginning of Section 4.5.

Lemma 4.6.1. Let p be an odd prime. Let a ∈ Q with a not a pth power in Q,and let E be the splitting field of X p − a. Let G = Gal(E/Q). Then

G = 〈σ, τ | σ p = 1, τ p−1 = 1, τστ−1 = σ r 〉for some primitive root r mod p.

Proof. Certainly E = Q(α, ζp) where α ∈ E, α p = a. We claim that Q(α) ∩Q(ζp) = Q. To see this, let B = Q(α) ∩ Q(ζp). Then Q ⊆ B ⊆ Q(ζp), so(B/Q) divides (Q(ζp)/Q) = p − 1, and Q ⊆ B ⊆ Q(α), so (B/Q) divides(Q(α)/Q) = p. Thus (B/Q) = 1, i.e., B = Q. In particular, α /∈ Q(ζp) and(E/Q(ζp)) = p. Thus, if mα(X) denotes the minimum polynomial of α overQ(ζp), and mα(X) denotes the minimum polynomial of α over Q, we see that

mα(X) = mα(X) =p−1∏i=0

(X − ζ ipα) ∈ E[X ].

Then, by either Lemma 2.6.1 or Lemma 2.6.3, there is an automorphism σ ofE with

σ(α) = ζpα, σ (ζp) = ζp.

4.6 Radical Polynomials and Related Topics 113

On the other hand, consider a generator τ0 of Gal(Q(ζp)/Q). This is givenby τ0(ζp) = ζ r

p for r a primitive root mod p. Again, by either Lemma 2.6.1 orLemma 2.6.3, there is an automorphism τ0 of E extending τ0. Then τ0(α) =ζ k

pα for some k. Set τ = σ−kτ0. Then τ is an automorphism of E with

τ(α) = α, τ(ζp) = ζ rp.

Then direct calculation shows

σ rτ(α) = σ r (τ (α)) = σ r (α) = ζ rpα

σ rτ(ζp) = σ r (τ (ζp)) = σ r (ζ rp) = ζ r

p ,

while

τσ (α) = τ(σ (α)) = τ(ζpα) = ζ rpα

τσ(ζp) = τ(σ (ζp)) = τ(ζp) = ζ rp ,

so σ rτ = τσ ∈ Gal(E/Q), i.e., τστ−1 = σ r , as claimed. ��Remark 4.6.2. (1) We see that, in the situation of Lemma 4.6.1, G = H N isthe semidirect product of the normal subgroup N = 〈σ | σ p = 1〉, isomorphicto Z/pZ, and the subgroup H = 〈τ | τ p−1 = 1〉, isomorphic to Z/(p − 1)Z.

(2) The structure of G is independent of the primitive root r . (There is achoice involved in the generators of G. Varying this choice varies r among allthe primitive roots.)

(3) Note that G is nonabelian. Note also that Lemma 4.6.1 also holds forp = 2, when it recovers the much easier fact that in this case G ∼= Z/2Z. �Lemma 4.6.3. Let a ∈ Q with a �= ±b2 for any b ∈ Q, and let E be thesplitting field of X4 − a. Let G = Gal(E/Q). Then

G = 〈σ, τ | σ 4 = 1, τ 2 = 1, τστ−1 = σ 3〉.Proof. Let α ∈ E with α4 = a. Then E = Q(α, ζ4). First we must showthat Q(α) ∩ Q(ζ4) = Q. Assume Q(α) ∩ Q(ζ4) ⊃ Q. Since (Q(ζ4)/Q) = 2,this implies that Q(α) ⊇ Q(ζ4) and hence that E = Q(α). Then (E/Q) = 4and (Q(ζ4)/Q) = 2, and so Q(ζ4) is the fixed field of a subgroup {id, σ } ofGal(E/Q). Since α is a root of the polynomial X4−a, σ(α) must also be a rootof this polynomial, so σ(α) = ζ k

4 α for some k. But σ(ζ4) = ζ4 and σ �= idbut σ 2 = id, so ζ k

4 α �= α but ζ 2k4 α = α. Hence k = 2 and σ(α) = −α. Now

E = {c0 + c1α + c2α2 + c3α

3 | ci ∈ Q} and Fix(σ ) = {c0 + c2α2 | ci ∈ Q}.

We claim that i = ζ4 /∈ Fix(σ ).Suppose i = c0 + c2α

2. Then −1 = i2 = (c0 + c2α2)2 = (c2

0 + c22a) +

2c0c2α2. Since a is not a square, α2 /∈ Q, so c0c2 = 0 and either c0 = 0,


c22 = −1/a, which is impossible since −a is not a square, or c2 = 0, c2

0 = −1,which is certainly impossible. Thus Q(α) ∩ Q(ζ4) = Q.

The rest of the proof entirely parallels the proof of Lemma 4.6.1. ��Lemma 4.6.1 and 4.6.3 have a considerable generalization in Theorem

4.6.6.We have seen in Section 4.5 that every quadratic field Q(

√a) is contained

in some cyclotomic field. We now use the theory we have developed to showthat no cubic, quartic, . . . field is contained in a cyclotomic field.

Corollary 4.6.4. Let a ∈ Q and let n > 2 be an integer. Suppose that a is nota pth power for some odd prime p dividing n or that ±a is not a square if n isdivisible by 4. Then Q( n

√a) is not contained in any cyclotomic field.

Proof. Let k = p or k = 4 according as n is divisible by the odd prime p orby 4.

Suppose that Q( n√

a) ⊆ Q(ζm). Since k√

a = ( n√

a)n/k , Q( k√

a) ⊆ Q( n√

a).Then E = Q( k

√a, ζk) ⊆ Q(ζm, ζk) ⊆ Q(ζmk). Now E is the splitting field of

Xk − a, so is a Galois extension of Q. Then, by the Fundamental Theorem ofGalois Theory,

Gal(E/Q) = Gal(Q(ζmk)/Q)/ Gal(Q(ζmk)/E).

In particular, Gal(E/Q) is a quotient of Gal(Q(ζmk)/Q). But we have thatGal(Q(ζmk)/Q) is abelian, by Corollary 4.2.7, while under our hypothesisGal(E/Q) is nonabelian, by Lemma 4.6.1 or Lemma 4.6.3, which is impos-sible. ��

We now prove a result which allows us to weaken the hypothesis of Lemma3.7.7 in our discussion of Kummer fields, as well as to further investigate theGalois groups of splitting fields of radical polynomials.

Lemma 4.6.5. (1) Let n be an odd integer, and let a be n-powerless in Q. Thena is n-powerless in Q(ζN ) for every N.

(2) Let n be an even integer, and let a be n-powerless in Q. If a is negative,assume also that −a is not a square in Q. Then one of the following twoalternatives holds:

(a) a is n-powerless in Q(ζN ) for every N.(b) a = b2 for some b ∈ Q(ζN ), for some N, and b is n/2-powerless in

Q(ζN ′) for every multiple N ′ of N.

Proof. If a is n-powerless in Q(ζN ), there is nothing to prove. Otherwise, a isan m th power in Q(ζN ) for some m dividing n, m > 1. We shall show that theonly possibility is m = 2, from which the lemma follows.


Suppose m > 2. Then m is divisible by k, where k is either an odd primeor 4. Write a = bk where b ∈ Q(ζN ). Then we see that Q( k

√a) is contained in

some cyclotomic field. But, under our hypothesis on a, this is impossible, byCorollary 4.6.4. ��Theorem 4.6.6. Let n be an integer and let a be n-powerless in Q. Let E bethe splitting field of Xn − a over Q, and let G = Gal(E/Q).

(1) If a is n-powerless in Q(ζn), then G = H N, the semidirect product ofa normal subgroup N isomorphic to Z/nZ and a subgroup H isomorphic to(Z/nZ)∗.

(2) If a = b2 with b n/2-powerless in Q(ζn), then G = H N, the semidirectproduct of a normal subgroup N isomorphic to Z/(n/2)Z and a subgroup Hisomorphic to (Z/nZ)∗.

Proof. (1) We have that E = Q(ζn, α), where αn = a. Let B = Q(ζn). ThenXn − a is irreducible over B, so by Proposition 3.7.6 the Galois group N =Gal(E/B) is isomorphic to Z/nZ. Now N is a subgroup of G and is normalas B is a Galois extension of Q. Also, H = Gal(E/Q(α)) is isomorphic toGal(B/Q) which is isomorphic to (Z/nZ)∗. Also, (E/Q) = (E/B)(B/Q) =(Q(α)/Q)(B/Q) so Q(α) and B are disjoint extensions of Q by Corollary3.4.5, so G = H N by Theorem 3.4.10.

(2) Let F = Q(b) ⊆ Q(ζn). Then (F/Q) = 2, and so G ′ = Gal(E/F) is asubgroup of index 2 of G. The proof of part (1) goes through with Q replacedby F and n replaced by m to give that G ′ = H ′N with H ′ = Gal(Q(ζn)/Q(b))

a subgroup of H = Gal(Q(ζn)/Q) of index 2. Now [G : G ′] = [H : H ′] = 2.To show G = H N it suffices to show that σ /∈ N where σ is any element of Hnot in H ′. But for any σ ∈ H , except σ = id, σ is non-trivial on B = Q(ζn),while every element of N acts trivially on B as N = Gal(E/B), so σ /∈ N , asrequired. ��Remark 4.6.7. As we have seen in our discussion of quadratic fields, both casesin Lemma 4.6.5 (2) may indeed arise. For example, 5 is 6-powerless in Q and√

5 /∈ Q(ζ6), so we are in case (a) for n = 6, while 5 is 10-powerless in Q but√5 ∈ Q(ζ10), so we are in case (b) for n = 10. �

We now wish to determine when the polynomial Xn − a is irreducible.Clearly if a is not n-powerless it is not, so we need only consider the casewhen a is n-powerless. Recall that in Corollary 4.1.10 we showed that Xn − ais irreducible if a is p-powerless for p an odd prime, or if a is positive and ais 2-powerless.

We shall generalize this result in several stages.


Corollary 4.6.8. Let a be n-powerless in Q. If n is odd, then the polynomialXn − a is irreducible in Q[X ].Proof. Since a is n-powerless in Q, and n is odd, we have, by Lemma 4.6.5,that a is n-powerless in Q(ζn). Then, by Proposition 3.7.6, Xn−a is irreduciblein Q(ζn)[X ], and hence Xn − a is certainly irreducible in Q[X ]. ��Lemma 4.6.9. Let n = 2t , and assume that a is not a square in Q. If t ≥ 2,assume also that −4a is not a 4th power in Q. Then the polynomial Xn − a isirreducible in Q[X ].Proof. Clearly the lemma holds when t = 1; so suppose t ≥ 2.

Let α be a root of X2t − a in some extension E of Q. Consider

Q = Q(a) = Q(α2t) ⊆ Q(α2t−1

) ⊆ · · · ⊆ Q(α).

Claim: (Q(α2k−1)/Q(α2k

)) = 2 for k = 1, . . . , t .Assuming this claim, we then immediately have that (Q(α)/Q) = 2t , and

hence that mα(X) is a polynomial of degree 2t . But α is a root of X2t − a, somα(X) = X2t − a and hence X2t − a is irreducible.

Proof of claim: We prove this by downward induction on k.If k = t , then α2k−1

is a root of X2−a, and a is not a square in Q = Q(α2k),

so (Q(α2k−1)/Q(α2k

)) = 2.Now suppose the claim is true for all integers between k and t and consider

k − 1. We wish to show (Q(α2k−2)/Q(α2k−1

)) = 2. We prove this by contradic-tion. Assume it is not the case. Then Q(α2k−2

) = Q(α2k−1). For simplicity, set

β = α2k−2, γ = α2k−1

, and δ = α2k. Now (Q(γ )/Q(δ)) = 2 by the inductive

hypothesis, so Q(γ ) has basis {1, γ } as a Q(δ)-vector space. Hence we maywrite β = c + dγ uniquely with c, d ∈ Q(δ). Then γ = β2 = (c + dγ )2 =(c2 + d2δ) + 2cdγ , so (c2 + d2δ) = 0 and 2cd = 1. Solving for δ yieldsδ = −4c4 with c ∈ Q(δ). We claim this is impossible.

If k − 1 = t − 1, then k = t and Q(δ) = Q. But then δ = −4c4 implies−4δ = 16c4 = (2c)4, contradicting the hypotheses of the lemma.

If k −1 < t −1, then k < t and (Q(δ)/Q(ε)) = 2 by the inductive hypoth-esis, where ε = α2k+1

. Thus, if ϕ is the nontrivial element of Gal(Q(δ)/Q(ε)),ϕ(δ) = −δ. Then −δ = ϕ(δ) = ϕ(−4c4) = −ϕ(4c4) = −ϕ((2c)2) =−(ϕ(2c))2, so δ = (ϕ(2c))2 with 2c ∈ Q(δ). But δ = γ 2, so γ = ±ϕ(2c) ∈Q(δ), contradicting the inductive hypothesis. ��Theorem 4.6.10. Let n be an integer and let a be n-powerless in Q. If n is di-visible by 4, assume also that −4a is not a 4th power in Q. Then the polynomialXn − a is irreducible in Q[X ].


Proof. Let n = n1n2 where n1 is odd and n2 is a power of 2. Let α be a rootof Xn − a in some extension field E of Q. Then αn2 is a root of Xn1 − a,which is irreducible by Lemma 4.6.9, so (Q(αn2)/Q) = n1 divides (Q(α)/Q).Similarly, αn1 is a root of Xn2 − a, which is irreducible by Lemma 4.6.10, so(Q(αn1)/Q) = n2 divides (Q(α)/Q). But n1 and n2 are relatively prime, son = n1n2 divides (Q(α)/Q). But certainly (Q(α)/Q) ≤ n, so (Q(α)/Q) = n,and Xn − a = mα(X) is irreducible. ��Remark 4.6.11. Theorem 4.6.10 is sharp. If n = 4m and −4a = b4, thenb = 2c and

Xn − a = (X2m + 2cXm + 2c2)(X2m − 2cXm + 2c2),

so Xn − a is not irreducible in this case. �Remark 4.6.12. (1) Theorem 4.6.10 and its proof show that the hypothesis inLemma 4.6.5 (2) may be weakened to: If a is a negative integer, assume alsothat −4a is not a 4th power in Z. However, it may not be weakened further. Forexample, let n = 4 and a = −4. Then a = (2i)2 in Q(ζ4) and 2i = (1 + i)2 inQ(ζ4).

(2) A similar observation applies to Theorem 4.6.6. If E is the splittingfield of X4 + 5, then | Gal(E/Q)| = 8 and we are in case (1) of that theorem.If E is the splitting field of X4 + 9, then | Gal(E/Q)| = 4 and we are in case(2) of that theorem. On the other hand, the roots of X4 + 4 are ±(1 ± i), so ifE is the splitting field of X4 + 4, then | Gal(E/Q)| = 2. �

We conclude this section by considering multiple radical extensions. Firstwe show they have the degrees we expect, and then we find primitive elements.Again we proceed in stages.

Theorem 4.6.13. Let {ni }, i = 1, . . . , t , be pairwise relatively prime inte-gers and let {ai }, i = 1, . . . , t be integers with ai ni -powerless for each i .If some ni is divisible by 4, assume that −4ai is not a 4th power in Z. ThenQ( n1

√a1, . . . , nt

√at ) is an extension of Q of degree n1 · · · nt .

Proof. Immediate from Theorem 4.6.12 and Lemma 3.4.2. ��Lemma 4.6.14. Let A = {a1, . . . , at } be a set of integers with the propertythat no product of the elements of any nonempty subset of A is a square.(In particular, this holds if a1, . . . , at are pairwise relatively prime, none ofwhich is a square, and at most one of which is the negative of a square.) ThenGal(Q(

√a1, . . . ,

√at )/Q) ∼= (Z/2Z)t . In particular, (Q(

√a1, . . . ,

√at )/Q)

= 2t .

First proof. This is a special case of Theorem 3.7.11.


Second proof. We proceed by induction on t . For t = 0 there is noth-ing to prove, and for t = 1 the result is obvious. By induction, it sufficesto prove that Q(

√a1, . . . ,

√at−1) and Q(

√at ) are disjoint extensions of Q.

Since (Q(√

at )/Q) = 2, to show this it suffices to show that Q(√

at ) �Q(

√a1, . . . ,

√at−1). Suppose

√at = x ∈ Q(

√a1, . . . ,

√at−1). Write x =

y + z√

at−1 with y, z ∈ Q(√

a1, . . . ,√

at−2). Squaring, at = (y2 + z2at−1) +2yz

√at−1. Since every term but the last is contained in Q(

√a1, . . . ,

√at−2),

by induction the last term must be zero, so either z = 0, in which casex ∈ Q(

√a1, . . . ,

√at−1), contradicting the inductive hypothesis, or y = 0,

in which case x = z√

at−1 and at = x2 = z2at−1 with z ∈ Q, which is cer-tainly impossible. ��

Theorem 4.6.15. Let n1, . . . , nt be integers and let a1, . . . , at be pairwise rel-atively prime integers with ai ni -powerless, i = 1, . . . , t . If some ni is even,assume that no ai is the negative of a square. Then

(Q( n1√

a1, . . . ,nt√

at )/Q) = n1 · · · nt

with basis {∏ti=1 aki /ni

i | 0 ≤ ki < ni }.Proof. Let n = lcm(n1, . . . , nt ). By hypothesis, each ai is ni -powerless. Wewill first prove the theorem under the stronger hypothesis that each ai is n-powerless, and then show how to remove this stronger hypothesis.

Thus, assume that each ai is n-powerless. Observe that in this case it suf-fices to prove the theorem for n1 = · · · = nt = n as clearly

(Q( n1√

a1, . . . ,nt√

at )/Q) ≤ n1 · · · nt

and

(Q( n√

a1, . . . ,n√

at )/Q( n1√

a1, . . . ,nt√

at )) ≤ (n/n1) · · · (n/nt ),

so

(Q( n√

a1, . . . ,n√

at )/Q) = nt

forces both inequalities to be equalities.

First, suppose n is odd. We prove the theorem by induction on n. It istrivially true for n = 1, so assume it is true for n′ < n and consider the casen′ = n.

Let F = Q(ζN ) where N is any multiple of n and let E = F( n√

a1, . . . ,n√

at ). (We need to consider not only N = n but also N a multiple of n forthe inductive step below.) Then, by Theorem 3.7.11, Gal(E/F) is isomorphic


to 〈a1, . . . , at 〉 ⊂ F∗/(F∗)n . Suppose a = ai11 · · · ait

t ∈ (F∗)n , i.e., a = αn forsome α ∈ F. In other words, a is not n-powerless in F, so, by the contraposi-tive of Lemma 4.6.5, a is not n-powerless in Q, and under our hypothesis ona1, . . . , at this is true if and only if each of i1, . . . , it is a multiple of n′′ forsome n′′ dividing n, n′′ > 1.

We claim that each i1, . . . , it is a multiple of n. If n′′ = n we are done, sosuppose not. Set n′ = n/n′′. Let a′ = ai1/n′′

1 · · · ait /n′′t . Then (a′)n′′ = a = αn =

(αn/n′′)n′′

, so a′ = ζ kn′′αn/n′′

for some k, i.e., a′ = (ζn/n′′n )kαn′

for α′ = ζ kn α, so

a′ ∈ (F∗)n′. Now n′ < n, so by induction each ik/n′′ is a multiple of n′ = n/n′′,

i.e., each ik is a multiple of n, as claimed.Since, by Lemma 2.3.4,

nt = | Gal(E/F)| = (E/F) ≤ (Q n√

a1, . . . ,n√

at )/Q) ≤ nt ,

we must have equality.

Next, suppose that n is even and let m = n/2. Choose N such thatQ(ζN ) ⊇ Q(ζn,

√a1, . . . ,

√at ), which is possible by Theorem 4.5.1, and set

F = Q(ζN ), E = F( n√

a1, . . . , n√

at ). Again, by Theorem 3.7.11, Gal(E/F) isisomorphic to 〈a1, . . . , at 〉 ⊂ F∗/(F∗)n , and now by the same argument as inthe case n odd,

mt = | Gal(E/F)| = (E/F) ≤ (Q( n√

a1 . . . , n√

at )/Q(√

a1, . . . ,√

at )) ≤ mt ,

and we have equality. But (Q(√

a1, . . . ,√

at )/Q) = 2t , by Lemma 4.6.14, so(Q( n

√a1 . . . , n

√at )/Q) = 2t mt = nt , completing the proof in this case.

Now let us see that we may in fact assume that the ai are all n-powerless.To do so we must use an argument from elementary number theory. Firstlet i = 1, and let a1 have prime factorization a1 = pe1

1 · · · pekk , and let

d1 = gcd(e1, . . . , ek). Since a1 is n1-powerless, gcd(d1, n1) = 1. Let n1 bethe product of the primes p that divide n but that do not divide n1. By theChinese Remainder Theorem, for each i there is an integer e′

i satisfying thecongruences

e′i ≡ ei (mod n1), e′

i ≡ 1(mod n1).

Let d = gcd(e′1, . . . , e′

k). We claim that gcd(d, n) = 1. To see this, con-sider a prime p dividing n. If p divides n1, then, since gcd(d1, n1) = 1, pdoes not divide ei for some i . But e′

i ≡ ei (mod n1) implies e′i ≡ ei (mod p),

so p does not divide e′i . If p does not divide n1, then p divides n1. But

e′i ≡ 1(mod n1) for every i , and e′

i ≡ 1(mod n1) implies e′i ≡ 1(mod p),

so p does not divide e′i .


Now let a′1 = p

e′1

1 · · · pe′

kk . Then a′

1 is n-powerless. Also, writing e′i = ei +

xi n1, note that

n1

√a′

1 = n1

√pe1+x1n1

1 · · · pet +xt n1t

= n1

√(pe1

1 · · · pekk )(px1

1 · · · pxtt )n1

= (px11 · · · pxt

t )n1

√(pe1

1 · · · pekk ) = r1

n1√

a1

where r1 = (px11 · · · pxt

t ) ∈ Q.Proceeding in exactly the same fashion, we may obtain a′

1, . . . , a′t with

each a′i n-powerless, and with ni

√a′

i = rini√

ai with ri ∈ Q for each i . Butthen Q( n1

√a′

1, . . . ,nt√

a′t ) = Q( n

√a1, . . . , n

√at ), so we may replace ai by a′

i andargue as above. ��Remark 4.6.16. (1) To illustrate a point in the proof of Theorem 4.6.15, leta1 = 8 = 23, n1 = 5, a2 = 3, n2 = 3. Then n = 15 and a1 is not 15-powerless,but we may rechoose a1 = 256 = 28 = 23+5 which is 15-powerless.

(2) Clearly Theorem 4.6.15 is true somewhat more generally, but the bestresult is clumsy to state. Our hypothesis were chosen so that we could directlyappeal to Lemma 4.6.5. �Corollary 4.6.17. Let n be an integer and let a1, . . . , at be pairwise relativelyprime n-powerless integers. If n is even, suppose that no ai is the negative ofa square and that n is relatively prime to each ai . Let E be the splitting field off (X) = ∏t

i=1(Xn −ai ) over Q. Then Gal(E/Q) is the semidirect product of agroup isomorphic to (Z/nZ)t and a normal subgroup isomorphic to (Z/nZ)∗.

Proof. Let B = Q( n√

a1, . . . , n√

at ) and let D = Q(ζn). Under our hypothe-ses, B and D are disjoint extensions of Q, by Corollary 4.6.4 or Corol-lary 4.5.5. Now E = BD and, by Theorem 4.6.15 and Corollary 3.4.4,| Gal(E/D)| = (E/D) = (BD/D) = (B/B ∩ D) = (B/Q) = nt . ClearlyGal(E/D) ⊆ (Z/nZ)t , as any automorphism must take n

√ai to ζ n

√ai for

some ζ with ζ n = 1, so these two groups are equal. Now Gal(D/Q) =Gal(Q(ζn)/Q) ∼= (Z/nZ)∗, so the result follows from Theorem 3.4.10. ��Lemma 4.6.18. Let {ni }, i = 1, . . . , t , be pairwise relatively prime integersand let {ai }, i = 1, . . . , t , be integers with ai ni -powerless for each i . If someni is divisible by 4, assume that −4ai is not a 4th power in Z. Then α =

n1√

a1 + · · · + nt√

at is a primitive element of Q( n1√

a1, . . . , nt√

at ).

Proof. Let N = n1 · · · nt and let Bi = Q( ni√

ai ), for each i . By Theorem4.6.13, (Bi/Q) = ni . Let B = B1 · · · Bt = Q( n1

√a1, . . . , nt

√at ). Let F =

Q(ζN ), let Di = FBi = F( ni√

ai ), for each i , and let D = D1 · · · Dt .


First, suppose each ni is odd. Then Bi and F are disjoint extensions of Qby Corollary 4.6.4, and F is a Galois extension of Q, so, by Corollary 3.4.4,(Di/F) = ni . Since the {ni } are pairwise relatively prime, we have, applyingLemma 3.4.2 successively, that D2 is disjoint from D1 (as extensions of F), andthen that D3 is disjoint from D1D2, . . . . Consequently, (D/F) = N . Clearlyαi = ni

√ai is a primitive element of Di as an extension of F. Then, applying

Proposition 3.5.5 successively, we have that α1 + α2 is a primitive element ofD1D2, . . . , and finally that α is a primitive element of D as an extension of F.It remains to show that α is a primitive element of B as an extension of Q. LetB′ = Q(α). Then, by Lemma 2.3.4,

N = (D/F) = (FB′/F) ≤ (B′/Q) ≤ (B/Q) = N ,

so we must have equality and B′ = B.Now suppose some ni , say n1, is even. The same argument as above

applies, except that (D1/F) = n1 or n1/2, enabling us to conclude that(D/F) = N or N/2 and hence that B = B′ or that (B/B′) = 2. But in thelatter case we must have B′ = Q( n1/2

√a1, n2

√a2, . . . , nt

√at ), which is impossi-

ble as α is not an element of this field. ��Lemma 4.6.19. Let n1, . . . , nt be integers and let a1, . . . , at be pairwise rel-atively prime integers with ai ni -powerless, i = 1, . . . , t . If some ni is even,assume that no ai is the negative of a square. Then α = n1

√a1 + · · · + nt

√at is

a primitive element of Q( n1√

a1, . . . , nt√

at ).

Proof. The proof proceeds along the same lines as the proof of Lemma 4.6.18.We need to know, in the notation of that proof, that (D/F) = N . In the easycase when n1 = · · · = nt = 2, F = Q and that follows immediately fromLemma 4.6.14. In the general case, we have shown that in the proof of The-orem 4.6.15. We then apply that fact to conclude that D2 is disjoint from D1

(as extensions of F), and then that D3 is disjoint from D1D2 . . . , (as otherwise(D/F) < N ). The remainder of the argument is the same. ��

We assemble these last two results into one theorem.

Theorem 4.6.20. Let n1, . . . , nt be integers and let a1, . . . , at be positive in-tegers with ai ni -powerless, i = 1, . . . , t . If ni and n j are not relatively prime,assume that ai and a j are relatively prime. Then α = n1

√a1 + · · · + nt

√at is a

primitive element of Q( n1√

a1, . . . , nt√

at ).

Proof. We proceed by induction on t . The t = 1 case is trivial.Now assume the theorem is true for all positive integers at most t . Consider

the case t+1. Let A = {ni | i = 1, . . . , t}, let A1 ={ni ∈ A | gcd(ni , nt ) = 1},


and let A2 = {ni ∈ A | gcd(ni , nt ) > 1}. If A1 = A, then this is just Lemma4.6.18, and if A2 = A, this is just Lemma 4.6.19. Thus we must consider the“mixed” case. Reordering if necessary, we may assume A1 = {n1, . . . , ns}and A2 = {ns+1, . . . , nt } for some s with 1 ≤ s < t . Then, by Lemma 4.6.18,α1 = n1

√a1 + · · · + ns

√as is a primitive element of B1 = Q( n1

√a1, . . . , ns

√as),

and, by Lemma 4.6.19, α2 = ns+1√

as+1, . . . , nt+1√

at+1 is a primitive element ofB2 = Q( ns+1

√as+1, . . . , nt+1

√at+1).

Furthermore, if F = Q(ζN ) for N = n1 · · · nt+1, B1F and B2F are dis-joint Galois extensions of F, so we may apply Proposition 3.5.5 as aboveto conclude that α = α1 + α2 is a primitive element of B1B2F; then wecontinue to argue as above to conclude that α is a primitive element ofB1B2 = Q( n1

√a1, . . . , nt+1

√at+1), completing the inductive step. ��

We conclude this section with an example that illustrates these last severalresults.

Example 4.6.21. (1) Let E = Q(√

2,3√

2,5√

2,7√

2). Then E is an extension ofQ of degree 210 with primitive element

√2 + 3

√2 + 5

√2 + 7

√2.

(2) Let E = Q(√

2,√

3,√

5,√

7). Then E is an extension of Q of de-gree 16 with primitive element

√2 + √

3 + √5 + √

7. Similarly, if E =Q(

3√

2,3√

3,3√

5,3√

7), then E is an extension of Q of degree 81 with primitiveelement 3

√2 + 3

√3 + 3

√5 + 3

√7.

(3) Let E = Q(√

2,3√

2,√

3,3√

3). Then E is an extension of Q of degree36 with primitive element

√2 + 3

√2 + √

3 + 3√

3. �

4.7 Galois Groups of Extensions of Q

We showed in Lemma 3.1.2 that every finite group is the Galois group of someGalois extension. It is natural to ask whether every finite group is the Galoisgroup of some Galois extension of Q. The answer to this question is unknown.

In this section we shall investigate two special cases: abelian extensions,i.e., extensions with abelian Galois group, and symmetric extensions, i.e., thesplitting fields of polynomials of degree n with Galois group the symmetricgroup Sn . (Abelian extension is standard terminology but symmetric extensionis not.)

Theorem 4.7.1. Every finite abelian group is the Galois group of a Galoisextension E of Q.

Proof. Let H be an abelian group and write H as the direct sum of cyclicgroups of prime power order, H = H1 ⊕ · · · ⊕ Hs . Let Hi have order mi .

4.7 Galois Groups of Extensions of Q 123

Choose distinct primes {pi } with pi ≡ 1(mod mi ), i = 1, . . . , s. Such primespi exist by Lemma C.1.1.

Let n0 = 1 and ni = p1 · · · pi for i > 0. Then Q(ζni−1) and Q(ζpi ) aredisjoint extensions of Q for each i > 0, by Corollary 4.2.8, so by Theorem3.4.7, applied inductively, we see that G = Gal(Q(ζns )/Q) = G1 ⊕ · · · ⊕ Gs

where Gi∼= Z/(pi − 1)Z. Since the order of Hi divides the order of Gi for

each i , we see that H is a quotient of G, so by the Fundamental Theoremof Galois Theory there is a field E intermediate between Q and Q(ζns ) withGal(E/Q) = H . ��Remark 4.7.2. Instead of using Lemma C.1.1, we may use a deep and famoustheorem of Dirichlet: Every arithmetic progression ax + b with a and b rel-atively prime contains infinitely many primes. However, we have chosen toinclude Lemma C.1.1 in order to give a self-contained and elementary proofof Theorem 4.7.1, rather than to simply quote this deep result. �Remark 4.7.3. We can ask: When can a Galois extension E of Q be a subfieldof a cyclotomic field Q(ζn) for some n? Since Gal(Q(ζn)/Q) is abelian andE ⊆ Q(ζn) implies Gal(E/Q) is a quotient of Gal(Q(ζn)/Q), an obvious nec-essary condition is that E be an abelian extension of Q, i.e., that Gal(E/Q) beabelian. This necessary condition turns out to be sufficient. This is the famousKronecker–Weber Theorem: Every abelian extension of Q is a subfield of acyclotomic field. The proof of this deep theorem is far beyond the confines ofthis book. (Note that we proved a very special case by direct construction. Weshowed in Theorem 4.5.1 that every quadratic extension of Q is a subfield ofa cyclotomic field. Of course, our construction in Theorem 4.7.1 produced asubfield of a cyclotomic field as well.) �

Now we turn to the case of symmetric extensions. First, we treat the caseof a polynomial of prime degree, which is relatively simple. Then we handlethe case of a polynomial of arbitrary degree, using a construction of van derWaerden.

Theorem 4.7.4. Let f (X) ∈ Q[X ] be an irreducible polynomial of degree p,p a prime, with exactly p − 2 real roots. Let E be the splitting field of f (X).Then Gal(E/Q) = Sp, the symmetric group on p elements.

Proof. Recall that since f (X) is irreducible and char(Q) = 0, f (X) musthave p distinct roots in C. Since f (X) ∈ R[X ], the two nonreal roots of f (X)

must be conjugates of each other.Let the roots of f (X) in C be α1, . . . , αp with α1 and α2 nonreal. Let τ be

complex conjugation. Then τ(α1) = α2, τ(α2) = α1, τ(αi ) = αi for i > 2.


We identify G = Gal(E/Q) with a subgroup of Sp by its action permuting{α1, . . . , αp}. Under this identification, τ is the transposition (1 2).

Now, since f (X) is irreducible, G acts transitively on {α1, . . . , αp}. Thuswe see that G is a transitive subgroup of Sp, p a prime, that contains a trans-position. Then, by Lemma A.3.1, G = Sp. ��Example 4.7.5. We now show that there exists a polynomial satisfying thehypotheses of Theorem 4.7.4, for each prime p. For p = 2 the result is trivial.Suppose now that p is an odd prime. Fix p and let

f (X) = X (X − 2)(X − 2a)(X − 4a) · · · (X − 2(p − 2)a) − (b + 2).

We will show that with the proper choice of a and b, f (X) is irreducibleand has exactly p − 2 real roots.

Let g(X) = X (X − 2)(X − 2a)(X − 4a) · · · (X − 2(p − 2)a), so g(X)

is a polynomial of degree p with roots 0, 2, 2a, 4a, . . . , 2(p − 2)a. Then,by elementary calculus, g′(X) = f ′(X), a polynomial of degree p − 1, hasroots s1, . . . , sp−1 with 0 < s1 < 2 < s2 < 2a < · · · < sp−1 < 2(p − 2)a.Then we see that f (X) is increasing on the interval (−∞, s1), decreasing onthe interval (s1, s2), increasing on the interval (s2, s3), . . . , and increasing onthe interval (sp−1, ∞). We also see that the maximum value of f (X) on theinterval [0, 2] is at most (2)(2)(2a)(4a) · · · (2(p − 2)a) = 2pa p−2(p − 2)!,and that g(3a) ≥ a p, g(7a) ≥ a p, g(11a) ≥ a p, and so forth. Thus we choose

a = 2p(p − 2)! and b = 2pa p−2(p − 2)!.First, with this choice of a and b, f (X) is irreducible by Eisenstein’s Cri-

terion (Proposition 4.1.7).Next, with this choice of a and b, f (X) < 0 for x ≤ 2, and f (3a) > 0,

f (7a) > 0, f (11a) > 0, and so forth. Now s1 < 2 < s2 < 2a < s3. Thusf (X) is decreasing on the interval [2, s2], increasing on the interval [s2, 2a],with f (2) < 0, f (2a) < 0, so f (X) has no roots on the interval [2, 2a].Then s2 < 2a < s3 < 4a < s4. Thus f (X) is increasing on the interval[2a, s3], decreasing on the interval [s3, 4a], with f (2a) < 0, f (4a) < 0, andf (s3) ≥ f (3a) > 0, so f (X) has exactly two roots r1 and r2 on the interval[2a, 4a]. By the same argument, f (X) has no roots on the interval [4a, 6a]and exactly two roots r3 and r4 on the interval [6a, 8a]. Proceeding in thisfashion we find roots r1, r2, . . . , rp−3 of f (X) on the interval (−∞, 2(p −2)].But sp−1 < 2(p − 2)a, so f (X) is increasing on the interval [2(p − 2)a, ∞),and f (2(p − 2)a) < 0, so there is exactly one more root rp−2 of f (X) in thatinterval, and lastly f (X) has a total of p − 2 real roots, as claimed. �


Now we turn to the case of arbitrary degree, which is considerably harder.Let f (X) ∈ F[X ] be a separable polynomial and let E be a splitting fieldof f (X). Let G = Gal(E/F). If the irreducible factors of f (X) are all dis-tinct, then f (X) has distinct roots, and eliminating repeated irreducible fac-tors will not change E, so we assume f (X) has distinct roots {α1, . . . , αn}.Then G permutes {α1, . . . , αn} (and this determines the action of G on E). Let{Y1, . . . , Yn} be a set of indeterminates and observe that the symmetric groupSn acts in a natural way as a group of permutations of this set. Then thereis an isomorphism G → G ′ ⊆ Sn given by σ �→ σ ′ where σ ′(Yi ) = Y j ifσ(αi ) = α j .

For brevity, let F denote the field of rational functions F = F(Y1, . . . , Yn)

and let E denote the field of rational functions E = E(Y1, . . . , Yn).

Definition 4.7.6. In the above situation, let

θ = α1Y1 + · · · + αnYn ∈ E

and let

F(Z) =∏τ ′∈Sn

(Z − τ ′(θ)) ∈ E[Z ]. �

The polynomial F(Z) is a symmetric function of its roots, so by Lemma3.1.12 its coefficients are functions of the elementary symmetric functions ofits roots and hence of Y1, . . . , Yn and the coefficients of f (X). Thus, F(Z) ∈F[Z ]. Now we may factor F(Z) into a product of irreducibles in F[Z ],

F(Z) = F1(Z) · · · Ft (Z).

One of these is divisible by Z − θ in E(Z). Renumbering if necessary we mayassume it is F1(Z). Since F(Z) is invariant under Sn , the action of Sn permutesthese factors.

Lemma 4.7.7. In the above situation, let H ′ ⊆ Sn be the subgroup leavingF1(Z) invariant, i.e.,

H ′ = {τ ′ ∈ Sn | τ ′(F1(Z)) = F1(Z)}.Then G ′ = H ′, and hence G is isomorphic to H ′.

Proof. We begin by making several observations.Note that, for any τ ′ ∈ Sn , τ ′(Z − θ) divides τ ′(F1(Z)), so

H ′ = {τ ′ ∈ Sn | τ ′(Z − θ) divides F1(Z)}.


Note also that G ×G ′ ⊆ G × Sn acts on E, and for any σ ∈ G, σσ ′(θ) = θ

(as σσ ′ simply permutes the terms of the sum α1Y1 + · · · + αnYn), and soσ(θ) = (σ ′)−1(θ).

Consider

G1(Z) =∏σ∈G

(Z − σ(θ)).

This is obviously invariant under G. Since the fix set of G operating on E isF, the fix set of G operating on E is F, by Corollary 3.4.9, so G1(Z) ∈ F[Z ].Since {α1, . . . , αn} is distinct, {σ(θ) | σ ∈ G} is distinct. Hence, by Lemma2.7.12, G1(Z) is irreducible in F[Z ]. Since G1(Z) has Z − θ as a factor inF[Z ], we must have that G1(Z) = F1(Z). Also, by Lemma 3.5.4, E = F(θ).Hence the action of Sn on E is determined by its action on θ .

Suppose τ ′ ∈ H ′. Then τ ′(Z − θ) is also a factor of F1[Z ] = G1[Z ], so itmust be Z −σ(θ) for some σ ∈ G. But τ ′(Z − θ) = Z − τ ′(θ) = Z −σ(θ) =Z − (σ ′)−1(θ), so τ ′ = (σ ′)−1. But σ ′ ∈ G ′ so τ ′ ∈ G ′.

On the other hand, if τ ′ /∈ H ′, then τ ′(Z − θ) = Z − τ ′(θ) is not a factorof F1(Z) = G1(Z), so it is not equal to Z − σ(θ) = Z − (σ ′)−1(θ) for anyσ ′ ∈ G ′. Thus τ ′ �= (σ ′)−1 for any σ ′ ∈ G ′ and τ ′ /∈ G ′. ��Remark 4.7.8. From the proof of Lemma 4.7.7, we see that

F1(Z) =∏

σ ′∈G ′(Z − σ ′(θ)),

and in fact

Fi (Z) =∏

τ ′∈τ ′i G ′

(Z − τ ′(θ))

where the product is taken over a left coset of G ′. In particular, since {α1, . . . ,

αn} are distinct, {τ ′(θ) | τ ′ ∈ Sn} are also all distinct. We see from this thateach Fi (Z) has distinct roots and no two Fi (Z) and Fj (Z) have a root incommon, for i �= j . Furthermore, we also see from this that the action of Sn

permutes {Fi (Z)} transitively. �Lemma 4.7.9. Let p be a prime. Let f (X) ∈ Z[X ] be a monic polynomial andlet f (X) be its image in Fp[X ] (identifying Fp with Z/pZ). Let E be a splittingfield of f (X) over Q and let E be a splitting field of f (X) over Fp. Supposethat all roots of f (X) in E are simple. Then K = Gal(E/Fp) is isomorphic toa subgroup of G = Gal(E/Q).


Proof. Begin with the polynomial f (X) and form F(Z) as in Definition 4.7.6.Since F(Z) is a polynomial in Y1, . . . , Yn, Z , its factorization in the rationalfunction field Q(Y1, . . . , Yn) is actually a factorization in the polynomial ringZ[Y1, . . . , Yn]. This follows as the coefficients of F(Z) are in fact polynomialsin the coefficients of f (X) by Lemma 3.1.14, and by the fact that Corollary4.3.8 generalizes from Z to Z[Y1, . . . , Yn] as Z[Y1, . . . , Yn] is a unique factor-ization domain. Thus we have

F(Z) = F1(Z) · · · Ft (Z)

with each Fi (Z) irreducible. Then reducing mod p gives a factorization (wherethe factors may not be irreducible)

F(Z) = F1(Z) · · · Ft (Z).

Let {Fi j (X)} be the irreducible factors of Fi (X), for i = 1, . . . , t . Let K ′be the subgroup of Sn consisting of those permutations in Sn that preserve theirreducible factor F11(X) of F1(X). Now the action of Sn permutes {Fi (X)}and hence {Fi j (X)}. By Remark 4.7.8, applied to E, the Fi (X) have no com-mon root and hence no common irreducible factor, so if F11(X) is preserved,F1(X) must be preserved. In other words, K ′ = {τ ′ ∈ Sn | τ ′(F11(X)) =F11(X)} ⊆ {τ ′ ∈ Sn | τ(F1(X)) = F1(X)} = H ′.

But by the same logic as before, K ′ is isomorphic to K = Gal(E/Fp). ��Proposition 4.7.10. For every positive integer n there exists a polynomialfn(X) ∈ Z[X ] of degree n whose splitting field E has Gal(E/Q) isomorphicto Sn.

Proof. Choose a monic polynomial f2(X) ∈ Z[X ] of degree n whose mod 2reduction f2(X) ∈ F2[X ] is irreducible, a monic polynomial f3(X) ∈ Z[X ]of degree n whose mod 3 reduction f3(X) ∈ F3[X ] is the product of an irre-ducible polynomial and a linear factor, and a monic polynomial f5(X) ∈ Z[X ]whose mod 5 reduction f5(X) ∈ F5[X ] is the product of an irreduciblequadratic and other factors that are distinct irreducible polynomials of odddegree. Let

f (X) = −15 f2(X) + 10 f3(X) + 6 f5(X).

Then f (X) is a monic polynomial that reduces mod p to f p(X) for p =2, 3, 5. Since f2(X) is irreducible, f (X) is irreducible, and hence its roots aredistinct.

Let G p be the Galois group of f p(X) over Fp, p = 2, 3, 5. By Lemma4.7.9, each G p is isomorphic to a subgroup of G = Gal(E/Q).


Since f2(X) is irreducible, G2 operates transitively on its roots, so G is atransitive subgroup of Sn .

Since f3(X) has an irreducible factor of degree n−1, and the Galois groupof a finite extension of a finite field is cyclic, G3 and hence G contains an(n − 1)-cycle.

By the same logic, G5 and hence G contains an element that is the productof a transposition and one or two cycles of odd order, so some (odd) power ofit is a transposition in G.

Hence, by Lemma A.3.2, G = Sn . ��Remark 4.7.11. Our work actually gives an algorithm for computing Galoisgroups of polynomials f (X) ∈ Q[X ]: Form the polynomial F(Z) as in Def-inition 4.7.6. Note that we do not need to know the roots of f (X) to do so,since, as we have observed, its coefficients are functions of Y1, . . . , Yn andthe coefficients of f (X). Then factor F(Z) into a product of irreduciblesF(Z) = F1(Z) · · · Ft (Z) in F[Z ]. Remark 4.1.13 gives an algorithm for doingso. As in Lemma 4.7.7, let H ′

i be the stabilizer of Fi (Z) in Sn . Note that, byRemark 4.7.8, the {H ′

i } are mutually conjugate subgroups of Sn so they areall isomorphic as abstract groups. (This is a general fact. If a group H actstransitively on a set S = {si } with Hi the stabilizer of si , then, if h ∈ H isany element with h(si ) = s j , we have that Hj = h Hi h−1.) Thus we see fromLemma 4.7.7 that, for any i , the Galois group of f (X) (i.e., the Galois groupGal(E/Q), where E is a splitting field of f (X)) is isomorphic to H ′

i . Thus wemay simply pick any factor Fi (X), and go through all of the (finitely many)elements of Sn , seeing which of them leave Fi (X) invariant, to obtain H ′

i andhence the Galois group.

It is evident that this algorithm is wholly impractical. We include it to showthat the problem of computing Galois groups is algorithmically solvable (notto give a practical method for their computation). �Corollary 4.7.12. For any finite group G, there is a field E that is a finiteextension of Q, and a subfield B of E (necessarily also a finite extension of Q),such that E is a Galois extension of B with Gal(E/B) ∼= G.

Proof. This follows immediately from Proposition 4.7.10 and the proof ofLemma 3.1.2. ��

4.8 The Discriminant

Let f (X) ∈ F[X ] be a polynomial of degree n, with splitting field E. As wehave seen, the Galois group Gal(E/F) is a subgroup of the symmetric group

4.8 The Discriminant 129

Sn . In this section we develop a criterion for deciding whether G is a subgroupof An , the alternating group.

Definition 4.8.1. Let f (X) ∈ F[X ] be a polynomial of degree n with rootsα1, . . . , αn in some splitting field E. Let

δ =∏i< j

(αi − α j )

and let

� = �( f (X)) = δ2.

� is the discriminant of f (X). �Remark 4.8.2. (1) Note that δ depends on the order of the roots α1, . . . , αn but�( f (X)) does not, so �( f (X)) is indeed an invariant of f (X).

(2) If f (X) is irreducible, then it has distinct roots, so in this case�( f (X)) �= 0.

(3) �( f (X)) is a symmetric polynomial in the roots of f (X), so, by Re-mark 3.1.5 and Lemma 3.1.14, it is a polynomial in the coefficients of f (X).Thus �( f (X)) ∈ F (and it is independent of the choice of splitting field E).

(4) �( f (X)) is visibly a square in E, but may or may not be a squarein F. �Lemma 4.8.3. Let f (X) ∈ F[X ] be an irreducible polynomial with splittingfield E and with Galois group G = Gal(E/F). Let � = �( f (X)) be thediscriminant of f (X).

(1) If � is a square in F, i.e., if δ ∈ F, then G ⊆ An.(2) If � is not a square in F, i.e., if δ /∈ F, then G � An.

Proof. We observe that E is a Galois extension of F so Fix(G) = F.Let σ ∈ Sn . Recall that σ ∈ An if and only if sign(σ ) = 1. Also,

sign(σ ) = (−1)i , where i is the number of inversions in σ , i.e., i is the numberof elements in the set I = {(i, j) | i < j but σ( j) < σ(i)}.

From this it follows immediately that σ(δ) = sign(σ )δ. Thus, if δ ∈ F,then σ(δ) = δ for every σ ∈ G, so G ⊆ An . On the other hand, if δ /∈ F, thenσ0(δ) �= δ for some σ0 ∈ G, and σ0 /∈ An , so G � An . ��Corollary 4.8.4. In the situation of Lemma 4.8.3, suppose δ /∈ F. Let B =F(δ). Also, let H = G ∩ An. Then E is a Galois extension of B with Galoisgroup Gal(E/B) = H, and B is a Galois extension of F with Galois groupGal(B/F) = G/H ∼= Z/2Z. ��


Corollary 4.8.5. Let f (X) ∈ F[X ] be an irreducible cubic with splitting fieldE, and let G = Gal(E/F). If �( f (X)) /∈ F, then G = S3, while if �( f (X)) ∈F, then G = A3

∼= Z/3Z.

Proof. By Theorem 2.8.19, G must be either S3 or A3, and then Lemma 4.8.3decides between the two. ��

In order to effectively use Lemma 4.8.3 we must be able to compute thediscriminant. We do that now, in low degrees.

Proposition 4.8.6. Let f (X) ∈ F[X ], and let � = �( f (X)).(1) If f (X) = X2 + a X + b, then � = a2 − 4b.(2a) If f (X) = X3 + bX + c, then � = −4b3 − 27c2.(2b) If f (X) = X3 + aX2 + bX + c, then � = −4a3c + a2b2 + 18abc −

4b3 − 27c2.

Proof. By Remark 4.8.2 (3), � is a polynomial in the elementary symmetricfunctions si of the roots of f (X), or, equivalently, in the coefficients of f (X).

(1) From Remark 3.1.5 we see that a = −s1 = −(α1 + α2) and thatb = s2 = α1α2. Directly from Definition 4.8.1 we see that � = (α1 − α2)

2 =α2

1 − 2α1α2 + α22 = α2

1 + 2α1α2 + α22 − 4α1α2 = a2 − 4b.

(2) From Remark 3.1.5 we see that a = −s1 = −(α1 + α2 + α3), thatb = s2 = α1α2 + α2α3 + α1α3, and that c = −s3 = −(α1α2α3). In principle,we can then simply find � by direct computation, but the computations getvery messy, so we look for a better way.

(2a) We observe that � is a homogenous polynomial of degree 6 in s1,s2, and s3, where si has degree i , or, equivalently, that � is a homogenouspolynomial of degree 6 in a, b, and c, where a has degree 1, b has degree 2, anda has degree 3. Since a = 0 in this case, � must be of the form � = sb3 + tc2

for some s and t . First, consider the polynomial f (X) = X3 − X with roots 1,0, and −1. Direct computation shows 4 = � = s(−1)3 + t (0)2, so s = −4.Next consider f (X) = X3 − 1 with roots 1, ω, and ω2. Direct computationshows −27 = � = s(0)2 + t (−1)2, so t = −27.

(2b) Note that adding the same value to each root leaves � unchanged, as itonly depends on the difference between roots. If f (X) = X3 +aX2 +bX + c,let X = Y − a/3. Then f (X) = g(Y ) = Y 3 + b′Y + c′ and so � = −4(b′)3 −27(c′)2. But, by elementary algebra, b′ and c′ can be expressed in terms of a,b and c, and doing so and substituting yields the result. ��Example 4.8.7. (1) Let f (X) = X3 + X + 1 ∈ Q[X ] and let f (X) have split-ting field E. Then f (X) is irreducible and �( f (X)) = −31 so Gal(E/Q) =S3. On the other hand, if B = Q[√−31], then Gal(E/B) = A3.

4.9 Practical Computation of Galois Groups 131

(2) Let f (X) = X3 − 3X + 1 ∈ Q[X ] and let f (X) have splitting field E.Then f (X) is irreducible and �( f (X)) = 81, so Gal(E/Q) = A3

∼= Z/3Z.Note we have already encountered this polynomial in Example 2.9.8. �

4.9 Practical Computation of Galois Groups

Our focus in this book has been the theoretical development of Galois theory,rather than its computational aspects. However, in the course of this develop-ment we have proved a number of results that are useful in (hand) computationof Galois groups of polynomials f (X) ∈ F[X ]. By this (common) languagewe mean the Galois groups Gal(E/F), where E is a splitting field of f (X).For the convenience of the reader, we shall collect most of these results in thissection. We shall keep the original numbering to make it easy for the reader torefer back to the place they originally appeared.

It is not only the individual results we have derived that are useful, but alsotheir interplay, and this better illustrated than summarized. Thus we concludethis section by working out a specific example that illustrates the applicationof many of these results together.

We begin by considering Galois groups of specific polynomials. Since weare in characteristic 0, the prime field F0 = Q and we may choose as primitiventh root of unity ζn = exp(2π i/n).

Recall we have the following definition:

Definition 3.7.5. Let α ∈ F. Then α is n-powerless in F if α is not anm th power in F for any m dividing n, m > 1.

In determining whether an element of F is n-powerless, the following resultsare useful:

Lemma 4.6.5. (1) Let n be an odd integer, and let a be n-powerlessin Q. Then a is a n-powerless in Q(ζN ) for every N.(2) Let n be an even integer, and let a be n-powerless in Q. If a isnegative, assume also that −a is not a square in Q. Then one of thefollowing two alternatives holds:(a) a is n-powerless in Q(ζN ) for every N.(b) a = b2 for some b ∈ Q(ζN ), for some N, and b is n/2-powerlessin Q(ζN ′) for every multiple N ′ of N.

Corollary 4.5.5. Let a be a square-free integer. If a is positive andodd, set a′ = a. Otherwise set a′ = 4|a|. Then a = b2 for someb ∈ Q(ζn) if and only if a′ divides n.


In terms of this definition, we have the following result:

Proposition 3.7.7. Let F ⊇ F0(ζn). Let E be an extension of F. Thefollowing are equivalent:(1) E is the splitting field of Xn − a ∈ F[X ], for some a that is n-powerless in F.(2) E is a Galois extension of F with Gal(E/F) ∼= Z/nZ.

This result has a generalization:

Theorem 3.7.11. Let F ⊇ F0(ζn) and let E be the splitting field of

f (X) = (Xn − a1) · · · (Xn − at )

with each ai �= 0. Then Gal(E/F) is isomorphic to 〈a1, . . . , at 〉 ⊆F∗/(F∗)n (i.e., to the subgroup of F∗/(F∗)n generated by a1, . . . , at ).

Now we turn to Galois groups of polynomials over Q itself.We observe that the nth cyclotomic polynomial n ∈ Q[X ], and the poly-

nomial Xn − 1 ∈ Q[X ], both have splitting field Q(ζn). Recall that (Z/nZ)∗denotes the multiplicative group of units in (Z/nZ), of order ϕ(n). Then wehave:

Corollary 4.2.7. Gal(Q(ζn)/Q) ∼= (Z/nZ)∗.

Now we have some results on extensions whose Galois groups are non-abelian.

Lemma 4.6.1. Let p be an odd prime. Let a ∈ Q with a not apth power in Q, and let E be the splitting field of X p − a. LetG = Gal(E/Q). Then

G = 〈σ, τ | σ p = 1, τ p−1 = 1, τστ−1 = σ r 〉for some primitive root r mod p.

Lemma 4.6.3. Let a ∈ Q with a �= ±b2 for any b ∈ Q, and let E bethe splitting field of X4 − a. Let G = Gal(E/Q). Then

G = 〈σ, τ | σ 4 = 1, τ 2 = 1, τστ−1 = σ 3〉.These two results have a common generalization:

Corollary 4.6.17. Let n be an integer and let a1, . . . , at be pairwiserelatively prime n-powerless integers. If n is even, suppose that no ai

is the negative of a square and that n is relatively prime to each ai .Let E be the splitting field of f (X) = ∏t

i=1(Xn − ai ) over Q. ThenGal(E/Q) is the semidirect product of a group isomorphic to (Z/nZ)t

and a normal subgroup isomorphic to (Z/nZ)∗.


Now we have two separate results that are useful in showing that Galoisgroups are large:

Theorem 4.7.4. Let f (X) ∈ Q[X ] be an irreducible polynomial ofdegree p, p a prime, with exactly p − 2 real roots. Let E be the split-ting field of f (X). Then Gal(E/Q) = Sp, the symmetric group on pelements.

Lemma 4.7.9. Let p be a prime. Let f (X) ∈ Z[X ] be a monic polyno-mial and let f (X) be its image in Fp[X ] (identifying Fp with Z/pZ).Suppose that all roots of both f (X) and f (X) are simple. Let E bethe splitting field of f (X) over Q and let E be the splitting field off (X) over Fp. Then K = Gal(E/Fp) is isomorphic to a subgroup ofG = Gal(E/Q).

We now return to general fields F.

For a polynomial f (X) ∈ F[X ] we defined the discriminant �( f (X)) inDefinition 4.8.1. We then have:

Lemma 4.8.3. Let f (X) ∈ F[X ] be an irreducible polynomial withsplitting field E and with Galois group G = Gal(E/F). Let � =�( f (X)) be the discriminant of f (X).(1) If � is a square in F, i.e., if δ ∈ F, then G ⊆ An.(2) If � is not a square in F, i.e., if δ /∈ F, then G � An.

In low degrees we have the following computation:

Proposition 4.8.6. Let f (X) ∈ F[X ], and let � = �( f (X)).(1) If f (X) = X2 + aX + b, then � = a2 − 4b.(2a) If f (X) = X3 + bX + c, then � = −4b3 − 27c2.(2b) If f (X) = X3 + aX2 + bX + c, then � = −4a3c + a2b2 +18abc − 4b3 − 27c2.

For irreducible cubics (and for quadratics, too, but there the result is trivial),the discriminant gives us the complete answer.

Corollary 4.8.5. Let f (X) ∈ F[X ] be an irreducible cubic with split-ting field E, and let G = Gal(E/F). If �( f (X)) /∈ F then G = S3

while if �( f (X)) ∈ F then G = A3∼= Z/3Z.

Now we turn to more general results which are useful in considering poly-nomials that are not irreducible.

Let f (X) ∈ F[X ] with f (X) = f1(X) f2(X). Let Bi be a splitting fieldof fi (X), i = 1, 2, with B1 and B2 both subfields of some field A. Let Gi =Gal(Bi/F), i = 1, 2, and let G = Gal(E/F), where E = B1B2 is a splitting


field of f (X). If B1 and B2 are disjoint extensions of F, we can apply thefollowing result.

Theorem 3.4.7. Let B1 and B2 be disjoint Galois extensions of F.Then E = B1B2 is a Galois extension of F with Gal(E/F) =Gal(B1/F) × Gal(B2/F).

In Section 3.4 we gave some criteria for determining when two extensionsare disjoint. But, in any case, we have the following generalization of Theorem3.4.7.

Corollary 3.4.8. Let B1 and B2 be finite Galois extensions of F andlet E = B1B2, B0 = B1 ∩ B2. Then E is a Galois extension of Fand Gal(E/F) = {(σ1, σ2) ∈ Gal(B1/F) × Gal(B2/F) | σ1 | B0 =σ2 | B0}.Finally, we have a result that enables us to compare Galois groups over

different fields.

Corollary 3.4.6 (Theorem on Natural Irrationalities). Let f (X) ∈F[X ] be a separable polynomial and let B ⊇ F. Let E be a splittingfield for f (X) over F. Then EB is a splitting field for f (X) over B andGal(BE/B) is isomorphic to Gal(E/E ∩ B), a subgroup of Gal(E/F),with the isomorphism given by σ �→ σ | E.

We now present a single, rather elaborate, example, which shows how,with cleverness, hard work, and the use of a number of the results we haveproved so far, we can determine the structure of a reasonably large Galoisgroup. (We advise the reader to write out the diagram of intermediate fieldsand their inclusions in order to follow the argument here.)

Example 4.9.1. Let α = 3√

1 + √3 and let E be the splitting field of mα(X) ∈

Q[X ]. We determine G = Gal(E/Q).We observe that α3 = 1 + √

3, so α3 − 1 = √3 and (α3 − 1)2 = 3 and

hence α6 −2α3 −2 = 0. Thus α is a root of the polynomial X6 −2X3 −2 = 0.This polynomial is irreducible by Eisenstein’s criterion, so we see mα(X) =X6 − 2X3 − 2. We can readily find all the roots of this polynomial. Certainly(ωα)3 = 1 + √

3 and (ω2α)3 = 1 + √3, so ωα and ω2α are also roots.

But we can also see that if β3 − 1 = −√3 then (β3 − 1)2 = 3 and hence

β6 − 2β3 − 2 = 0, so β is also a root of mα(X) then ωβ and ω2β are roots aswell. Thus

E = Q(α, ωα, ω2α, β, ωβ, ω2β) = Q(ω, α, β).


Clearly E contains the fields Q(√

3), Q(ω) = Q(√−3), and Q(i) as well,

as i = (√

3)(√−3)/3. These are all extensions of Q of degree 2, and so we see

that G contains at least 3 subgroups of index 2. Since mα(X) is an irreduciblepolynomial of degree 6, (Q(α)/Q) = 6, so (Q(α)/Q(

√3)) = 3, and similarly

(Q(β)/Q(√

3)) = 3. Since Q(ω) and Q(√

3) are disjoint Galois extensionsof Q, both of degree 2, Q(ω,

√3) is a Galois extension of Q of degree 4,

and Q(ω,√

3) is a Galois extension of Q(√

3) of degree 2. Since 2 and 3 arerelatively prime, Q(ω,

√3) and Q(α) are disjoint extensions of Q(

√3), and

so Q(ω, α) = Q(ω, α,√

3) is an extension of Q(√

3) of degree 6. In fact, it isa Galois extension as it is the splitting field of the polynomial X3 − (1 + √

3).Similarly Q(ω, β) is a Galois extension of Q(

√3) of degree 6.

We now make an important observation. We have just noted that Q(ω,√

3)

and Q(α) are disjoint extensions of Q(√

3), so in particular ω /∈ Q(α). ThusQ(α) contains the root α of mα(X) but does not contain the root ωα, so Q(α) isnot a normal extension of Q(

√3). Hence we see that Gal(Q(ω, α)/(Q(

√3))

is a nonabelian group of order 6, as is Gal(Q(ω, β)/(Q(√

3)). Furthermore,since these two Galois groups are quotients of subgroups of G, we concludethat G is not abelian.

We claim β /∈ Q(ω, α). For if β ∈ Q(ω, α), we would have E = Q(ω, α)

and then (E/Q) = (Q(ω, α)/Q(√

3))(Q(√

3)/Q) = (6)(2) = 12, so G wouldbe a nonabelian group of order 12 with at least 3 subgroups of order 6, and nosuch group exists. Similarly, α /∈ Q(ω, β).

Now 6 = (Q(ω, α)/Q(√

3)) = (Q(ω, α)/Q(ω,√

3))(Q(ω,√

3)/Q(√

3)),so we see that (Q(ω, α)/Q(ω,

√3)) = (Q(ω, α)/Q(ω)) = 3, and similarly

we see that (Q(ω, β)/Q(ω)) = 3.

We claim that Q(ω, α) and Q(ω, β) are disjoint extensions of Q(ω,√

3).For we have the inclusions Q(ω,

√3) ⊆ Q(ω, α) ∩ Q(ω, β) ⊆ Q(ω, α), so

(Q(ω, α) ∩ Q(ω, β)/Q(ω,√

3)) divides (Q(ω, α)/Q(ω,√

3)) = 3, so it iseither 1 or 3, i.e., Q(ω, α)∩ Q(ω, β) = Q(ω, α) or Q(ω,

√3). But Q(ω, α)∩

Q(ω, β) �= Q(ω, α) as α /∈ Q(ω, β).

Since Q(ω, α) and Q(ω, β) are disjoint Galois extensions of Q(ω,√

3),we have that

(Q(ω, α, β)/Q(ω,√

3)) = 9 and hence (E/Q) = 36.

We also know that

9 = (Q(ω, α, β)/Q(ω,√

3))

≤ (Q(α, β)/Q(√

3))

≤ (Q(α)/Q(√

3))(Q(β)/Q(√

3)) = (3)(3) = 9,


so we see that (Q(α, β)/Q(√

3)) = 9. Hence (Q(α, β)/Q) = 18 and(E/Q(α, β)) = 2. Since

36 = (E/Q) = (Q(ω, α, β)/Q)

= (Q(ω)/Q)(Q(α, β)/Q) = (2)(18),

we see that Q(ω) and Q(α, β) are disjoint extensions of Q.

We now assemble all of this information to find the structure of G. First,since (E/Q) = 36, G is a group of order 36.

As we have observed, Gal(Q(ω, α)/Q(√

3)) is a nonabelian group of order6. By the Theorem on Natural Irrationalities, Gal(E/Q(β)) is isomorphic to asubgroup of this group. But (E/Q(β)) = 6 so these two groups are isomor-phic. In other words, every automorphism of Q(ω, α) fixing Q(

√3) extends

to an automorphism of E fixing Q(β), and similarly every automorphism ofQ(ω, β) fixing Q(

√3) extends to an automorphism of E fixing Q(α). Simi-

larly, since (E/Q(α, β)) = (Q(ω)/Q) = 2, we may again apply the Theoremon Natural Irrationalities to extend the nontrivial element of Gal(Q(ω)/Q) toan automorphism of E fixing Q(α, β).

The elements we have found so far generate a subgroup H of G of order18. Finally, although Q(α, β) is not a Galois extension of Q, E is a splittingfield of mα(X), an irreducible polynomial two of whose roots are α and β, sothere is an automorphism of E taking α to β. This automorphism is not unique,but if we choose one such, we may multiply it by any element of H . (Noteevery element of H leaves each of the sets {α, ωα, ω2α} and {β, ωβ, ω2β}invariant while this new automorphism interchanges these two sets.)

Thus, assembling all of this information, we find that G is a group of order36, generated by an element ρ of order 2, two elements σ1 and σ2 of order 3,and an element τ of order 2, whose actions on E are given by:

ρ(ω) = ω2, ρ(α) = α, ρ(β) = β

σ1(ω) = ω, σ1(α) = ωα, σ1(β) = β

σ2(ω) = ω, σ2(α) = α, σ2(β) = ωβ

τ(ω) = ω, τ(α) = β, τ(β) = α.

(Actually, H is generated by ρ, σ1, and σ2, and G is generated by ρ, σ1,and τ , as σ2 = ρσ1ρ

−1, but the more symmetric set of generators better revealsthe structure of G.)

Finally, if we number the roots α, ωα, ω2α, β, ωβ, ω2β of mα(X) as 1, . . . ,6, the generators of G are given by the permutations

ρ = (23)(56), σ1 = (123), σ2 = (456), and τ = (14)(25)(36). �

4.10 Exercises 137

4.10 Exercises

Exercise 4.10.1. Show, without using Dirichlet’s theorem about the existenceof infinitely many primes in arithmetic progressions, and without using LemmaC.1.1, that:(a) For any integer n, there exists a Galois extension E of Q of degree n.(b) For any integer n, and any algebraic number field F, there exists a Galoisextension E of F of degree n.

Exercise 4.10.2. For a rational number x = r/s in lowest terms, say that pdivides x if p divides r . Show that Eisenstein’s Criterion (Proposition 4.1.7)remains valid for polynomials in Q[X ] with this definition of divisibility.

Exercise 4.10.3. Let f (X) = an Xn + · · · + a0 ∈ Z[X ] and let r ∈ Q withf (r) = 0. Show that each of the n rational numbers

anr, anr2 + an−1r, anr3 + an−1r2 + an−2r,

. . . , anrn + an−1rn−1 + · · · + a1r

is in fact an integer. (This exercise is taken from the 2004 Putnam competition.)

Exercise 4.10.4. Use Exercise 2.10.10 to show that the polynomial X6 − 72 isirreducible in Q[X ].Exercise 4.10.5. We might call an ordinary compass a 2-compass since, alongwith a straightedge, it allows us to geometrically find square roots of complexnumbers. Suppose we also had a 3-compass, an instrument that allows us togeometrically find cube roots of complex numbers. For what values of n <

1000 could we construct a regular n-gon with these instruments?

Exercise 4.10.6. (a) Let n be an integer and let k be the product of the distinctprime factors of n. Show that n(X) = k(Xn/k).(b) Let n be an integer not divisible by p. Show that

np(X) = n(X p)/n(X).

Exercise 4.10.7. (a) Find the n-the cyclotomic polynomial n(X) for n = pk

a prime power.(b) Find n(X) for 2 ≤ n ≤ 16.(c) Examining n(X) for small values of n might lead one to conjecture thatthe coefficients of n(X) are always 0, 1, or −1. Show, by hand computation,that n = 105 is a counterexample to this conjecture. (This is known to be the


smallest counterexample.) Note that 105(X) is a polynomial of degree 48.With enough patience, this entire polynomial can be computed by hand. Butonly a partial, and much more manageable, computation is necessary to find acoefficient that is not 0, 1, or −1.

Exercise 4.10.8. Fix n > 2 and let E = Q(ζn) be the n-th cyclotomic field.Let B = E ∩ R.(a) Show that (E/B) = 2 and that B = Q(ζn + ζ−1

n ).(b) Show that B is the unique subfield of E with (E/B) = 2 if and only ifn = 4, pk, or 2pk for p an odd prime.(c) Find 3 subfields B of E with (E/B) = 2 in case n = 8.(d) Show that there are exactly 3 fields B with (E/B) = 2 if n = 2k for anyk ≥ 3.

Exercise 4.10.9. Let E = Q(ζp) be the p-th cyclotomic field. Let a, b ∈ Q bearbitrary. Find NE/Q(a + bζp) and TrE/Q(a + bζp).

Exercise 4.10.10. (a) Of course, ζ1 = 1, ζ2 = −1, ζ3 = ω = (−1 + i√

3)/2,and ζ4 = i . Find ζ5, ζ6, ζ8, ζ10, ζ12, ζ15, and ζ16.(b) Find ζ17. (This is an intricate computation. See Example 4.4.6 (3).)(c) Find ζ7, ζ9, ζ13, and ζ14. (This uses the solution of the cubic in Exercise4.10.25.)

Exercise 4.10.11. Throughout this exercise, we assume n > 2.(a) Show that n(x) > 0 for all real numbers x .(b) If n = pk for some k, show that n(1) = p.(c) If n is not a prime power, show that n(1) = 1.(d) If n = 2k for some k, show that n(−1) = 2.(e) If n = 2pk for some odd prime p, show that n(−1) = p.(f) If n is not a power of 2 and is not twice an odd prime power, show thatn(−1) = 1.(g) Let E = Q(ζn). Show that the computations in (b) and (c) give the valueof NE/Q(1 − ζn) and that the computations in (d), (e), and (f) give the value ofNE/Q(−1 − ζn).

Exercise 4.10.12. (a) Let F = Fp be the field with p elements and let let m bean integer relatively prime to p. Let m(X) ∈ F[X ] be the mod p reductionof the cyclotomic polynomial m(X) and let g(X) = Xm − 1 ∈ F[X ]. Showthat m(X) and g(X) have the same splitting field E, and furthermore thatE = Fpr where r is the smallest positive integer such that pr ≡ 1(mod m).(b) Let m(X) = f1(X) · · · fk(X) ∈ F[X ] be a factorization of m(X) intoa product of irreducible polynomials. Show that each fi (X) has degree r , andhence that k = ϕ(m)/r .

4.10 Exercises 139

Exercise 4.10.13. Referring to Exercise 3.9.24 (b), let k(Y, Z) = Y Z . Letm and n be arbitrary positive integers and let d = gcd(m, n) and � =lcm(m, n). Show that QY Z (m(X), n(X)) = QY Z (�(X), d(X)). In par-ticular, if m and n are relatively prime, show that QY Z (m(X), n(X)) =QY Z (mn(X), 1(X)) = mn(X).

Exercise 4.10.14. Factor each of the following polynomials into a product ofirreducible polynomials in Q[X ]. (This includes the possibility that the givenpolynomial is irreducible.)(a) f (X) = X3 − 7X2 + 36.(b) f (X) = X3 + 5X2 − 10X − 8.(c) f (X) = X3 − 4X2 − 5X + 6.(d) f (X) = X4 + 2X2 + 9.(e) f (X) = X4 + 7X3 + 17X2 + 18X + 6.(f) f (X) = X4 − 8X2 + 15.(g) f (X) = X4 − 4X2 + 16.(h) f (X) = X4 + X3 + 7X2 + 15X + 15.(i) f (X) = X5 + X4 + 4X3 + 4X2 + 4X + 4.(j) f (X) = X6 − 2X3 − 1.(k) f (X) = X7 − 7X6 + 6X + 30.

Exercise 4.10.15. For each α in Exercise 2.10.1, find mα(X) ∈ Q[X ]. (Con-sider the polynomial you found in solving that problem. If it is irreducible, it ismα(X). Thus, you must decide whether it is irreducible. If it is, you are done.If not, you have more work to do.)

Exercise 4.10.16. For each α in Exercise 4.10.15, let E be a splitting field ofmα(X). Find (E/Q) and find G = Gal(E/Q).

Exercise 4.10.17. In this exercise, let E be a splitting field of the polynomialf (X) ∈ Q[X ] and let G = Gal(E/Q).(1) Find (E/Q) and find G.(2) Find all subgroups H of G, and the fields B to which they belong.(3) For each field B in (b), find a basis for E as a vector space over B and abasis for B as a vector space over Q.(4) Determine which subfields B in (b) are Galois extensions of Q. For eachof these fields B, find a polynomial g(X) ∈ Q[X ] with B a splitting field forg(X), and find Gal(B/Q).(a) f (X) = X4 − 2.(b) f (X) = X5 − 2.(c) f (X) = (X3 − 2)(X3 − 3).(d) f (X) = (X2 − 3)(X3 − 1).


(e) f (X) = (X2 + 3)(X3 − 1).(f) f (X) = X6 + 3.(g) f (X) = X6 + 5.(h) f (X) = X8 − 2.(i) f (X) = X8 − 3.

Exercise 4.10.18. (a) Let f (X) = X3 + bX + c ∈ Q[X ] be irreducible. Sup-pose b > 0. Show that the Galois group of f (X) is isomorphic to S3.(b) More generally, let f (X) = X3 + aX2 + bX + c ∈ Q[X ] be irreducible.Suppose b > a2/3. Show that the Galois group of f (X) is isomorphic to S3.

Exercise 4.10.19. (a) Let p be an odd prime and let

f (X) = X2(X − 2)(X − 4) · · · (X − 2(p − 2)) − 2.

Show that, for k sufficiently large, f (X) is an irreducible polynomial withp − 2 real roots.(b) Let p be an odd prime and let

f (X) = (X2 + 2)(X)(X − 2) · · · (X − 2(p − 3)) + 2/(2k + 1).

Show that, for k sufficiently large, f (X) is an irreducible polynomial withp − 2 real roots. (Then, choosing 2k + 1 to be a pth power, i.e., 2k + 1 = s p

for some integer s, we may make the substitution Y = s X to obtain a monicirreducible polynomial with integer coefficients g(Y ) = (2k + 1) f (X) =s p f (X).)(c) Let p be an odd prime and let

f (X) = (X2 + 2k)(X)(X − 2) · · · (X − 2(p − 3)) + 2.

Show that, for k sufficiently large, f (X) is an irreducible polynomial withp − 2 real roots.

Hence, in any of these three cases, f (X) has Galois group Sp. CompareTheorem 4.7.4 and Example 4.7.5.

Exercise 4.10.20. Let f (X) ∈ Q[X ] be a reducible cubic. Show that f (X)

has splitting field Q(δ), where δ is as in Definition 4.8.1.

Exercise 4.10.21. Let α = 3√

7 + 5√

2. Then certainly (α3 − 7)2 = 50, so wesee that α is a root of f (X) = 0, where f (X) = X6 − 14X3 − 1 ∈ Q[X ]. Ob-serve that α = β3, where β = 1+√

2. Use this observation to find all the rootsof f (X), the factorization of f (X) as a product of irreducible polynomials inQ[X ], the splitting field of f (X), and the Galois group of f (X).

4.10 Exercises 141

Exercise 4.10.22. Let f (X) = X5 −75X4 +15X3 −9X2 −16X −56 ∈ Q[X ].Show that the Galois group of f (X) is S5.

Exercise 4.10.23. Let f (X) = 4X5 − 105X4 + 840X3 − 2160X2 + 6000 ∈Q[X ]. Show that the Galois group of f (X) is S5.

Exercise 4.10.24. Let G be a finite group of order n and suppose G can bewritten as G = H1×· · ·×Hk with Hi a group of prime power order, for each i .Such a group is called nilpotent. (This is not the definition of a nilpotent groupbut rather a characterization of finite nilpotent groups.)(a) Show that G is solvable.(b) Let E be an extension of F with Gal(E/F) ∼= G. Show that, for everydivisor d of n, there is a field B with F ⊆ B ⊆ E and with (B/F) = d.(c) Find a counterexample to (b) for G a solvable group.

Exercise 4.10.25. In this exercise we will show how to find the roots of a gen-eral cubic polynomial f (X). Clearly it suffices to consider the case that f (X)

is monic, so let f (X) = X3 + aX2 + bX + c. By making the substitutionY = X − a/3, we may transform f (X) to a cubic g(Y ) = Y 3 + b′Y + c′,i.e., to one in which the quadratic term is absent. Thus it suffices to handle thiscase, so we assume f (X) = X3 + bX + c. Recall from Section 4.8 that f (X)

has discriminant � = −4b3 − 27c2. Let E be the splitting field of f (X) andlet D = E(ω). Then (E/Q(

√�)) = 1 or 3 so we see from Section 3.7 that

if B = Q(ω,√

�), then either D = B or that D = B(ε) with ε3 = e ∈ B,i.e., that D is the splitting field of the polynomial X3 − e ∈ B[X ]. We shallnot try to find ε or e directly, but these considerations guide the form of ouranswer. Indeed, with these in mind, and the observation that e must be of theform e = u + v

√� for some u, v ∈ Q(ω), we look for roots of f (X) of the

form

α = 3

√u + v

√� + 3

√u − v

√�,

β = ω3

√u + v

√� + ω2 3

√u − v

√�,

γ = ω2 3

√u + v

√� + ω

3

√u − v

√�.

(Observe that α, β, and γ are all fixed by Gal(B/Q), as we expect.) Nowrecall that the coefficients of f (X) are, up to sign, the elementary symmetricfunctions in the roots of f (X). Since 1+ω+ω2 = 0, we have that α+β+γ = 0,which is consistent with the quadratic term of f (X) being absent. Examiningthe linear and constant terms we obtain the equations

αβ + βγ + αγ = b, αβγ = −c.


Show that these equations yield

u = −c/2, v = √−3/18.

and verify by direct substitution that, with these values of u and v, α, β, and γ

are roots of f (X) = X3 + aX + b.

Exercise 4.10.26. Use the formula in Exercise 4.10.25 to find the roots of thefollowing cubics:(a) f (X) = X3 − 3X − 1.(b) f (X) = X3 − 21X − 37.(c) f (X) = X3 − 12X + 18.(d) f (X) = X3 − 27X + 30.(e) f (X) = X3 + 3X − 4.(f) f (X) = X3 − 13X − 12.

Exercise 4.10.27. Observe that the polynomials in Exercise 4.10.26 parts (e)and (f) are not irreducible. Find their roots directly. Compare your answershere with your answers in Exercise 4.10.26. You may be surprised!

Exercise 4.10.28. (a) Let F ⊆ R and let E = F(α) where αn = a ∈ Fand α ∈ R. Let B be any Galois extension of F with B ⊆ E. Show that(B/F) = 1 or 2.(b) Use part (a) to show that the solution of a cubic must involve nonrealradicals even when there are only real roots. (Examples of this are provided byparts (a) and (d) of Exercise 4.10.26.)

5

Further Topics in Field Theory

5.1 Separable and Inseparable Extensions

We now wish to further investigate questions related to separability and insep-arability of algebraic extensions. Recall from Corollary 3.2.3 that every alge-braic extension in characteristic 0 is separable, so in this case there is nothingmore to be said. Thus in this section we shall assume that char(F) = p > 0.

Recall we have the Frobenius endomorphism : F → F given by (x) =x p, and Fp = Im() = {y ∈ F | y = x p for some x ∈ F} is a subfield of F.(See Definition 2.1.10 and Corollary 2.1.11.) If F is finite, then Fp = F, but inthis case, too, every algebraic extension of F is separable (Theorem 3.2.6), sowe will be interested in the case that F is infinite.

First, we shall deal with extensions that are separable, and then with ex-tensions that are not.

We begin with a technical lemma.

Lemma 5.1.1. Let E be an extension of F and let α ∈ E, α /∈ F. Then α p /∈ Fp.

Proof. Suppose α p ∈ Fp, i.e., α p = β p for some β ∈ F. Then 0 = α p −β p =(α − β)p so α = β. But α /∈ F and β ∈ F, so this is impossible. ��

We also remind the reader of the following result.

Lemma 5.1.2. Let a ∈ F, a /∈ Fp. Then f (X) = X pt −a ∈ F[X ] is irreduciblefor every t ≥ 1.

Proof. Lemma 3.2.8. ��Lemma 5.1.3. Let E be a finite extension of F. If FEp = E, then E is a sepa-rable extension of F.


144 5 Further Topics in Field Theory

Proof. We claim that if {αi } is a set of elements of E that is linearly indepen-dent over F, then {α p

i } is also linearly independent over F. To see this, notethat, extending {αi } if necessary, we may assume that {αi }i=1,...,d , d = (E/F),is a basis for E over F. We will show that {α p

i }i=1,...,d spans E over F. Hence,since (E/F) = d is finite, {α p

i }i=1,...,d is a basis for E over F and hence this setis linearly independent.

Let γ ∈ E. Since E = FEp, we may write

γ = bα p for some b ∈ F, α ∈ E.

Since {αi } is a basis,

α = c1α1 + · · · + cdαd with c1, . . . , cd ∈ F,

and then

γ = bα p = b(c1α1 · · · + cdαd)p = (bcp

1 )αp1 + · · · + (bcp

d )αpd ,

proving the claim.Now suppose α ∈ E is not separable. Then, by Proposition 3.2.2, its mini-

mum polynomial mα(X) is of the form

mα(X) =k∑

i=0

ci Xip for some k and {ci }, not all zero.

Thus

0 =k∑

i=0

ciαi p,

so 1 = α0, α p, . . . , αkp are linearly dependent and hence, by the contrapositiveof our claim, 1 = α0, α, . . . , αk are linearly dependent as well.

Thus

0 =k∑

i=0

diαi with not all di equal to zero,

so if

f (X) =k∑

i=0

di Xi ,

then f (α) = 0. But deg f (X) < deg mα(X), which is impossible. ��

5.1 Separable and Inseparable Extensions 145

Lemma 5.1.4. Let α ∈ E. Then α is separable over F if and only if F(α) =F(α p).

Proof. Suppose that α is separable over F. Clearly F(α p) ⊆ F(α), so weneed only show the reverse inclusion. To show that, it suffices to show thatα ∈ F(α p).

Let mα(X) ∈ F[X ] be the minimum polynomial of α over F, and letmα(X) ∈ F(α p)[X ] be the minimum polynomial of α over F(α p). Thenmα(X) divides mα(X) in F(α p)[X ].

Clearly α is a root of the polynomial f (X) = X p − α p ∈ F(α p)[X ].Suppose α /∈ F(α p). Then α p /∈ (F(α p))p by Lemma 5.1.1, so f (X) = X p −α p is irreducible in F(α p)[X ] by Lemma 5.1.2. Since f (X) is irreducible,mα(X) = X p − α p. Since mα(X) divides mα(X) in F(α p)[X ], mα(X) dividesmα(X) in F(α)[X ]. Hence mα(X) is divisible by X p − α p = (X − α)p inF(α)[X ], contradicting the separability of α.

Conversely, if F(α) = F(α p), then

F(α) = F(α p) = F(F(α))p

so, by Lemma 5.1.3, F(α) is a separable extension of F, i.e., α is a separableelement of E. ��Corollary 5.1.5. If E is obtained from F by adjoining a root of a separablepolynomial, then E is a separable extension of F.

Proof. Let E = F(α) where α is a root of a separable polynomial. ThenF(α) = F(α p) by Lemma 5.1.4, and F(α p) = F(F(α))p, so F(α) = F(F(α))p

and F(α) is a separable extension of F by Lemma 5.1.3. ��Lemma 5.1.6. Let B be a field intermediate between F and E. If E is a sepa-rable extension of F, then B is a separable extension of F and E is a separableextension of B.

Proof. As B ⊆ E, B is trivially a separable extension of F.Let α ∈ E. Since E is a separable extension of F, α is a simple root of a

polynomial f (X) ∈ F[X ]. But F ⊆ B so f (X) ∈ B[X ] and α is separableover B. ��

Lemma 5.1.3 has a converse (under weaker hypotheses, in fact).

Lemma 5.1.7. Let E be an extension of F. If E is a separable extension of F,then FEp = E.


Proof. Let B = FEp so that B is intermediate between F and E. By Lemma5.1.6, E is a separable extension of B.

Suppose B ⊂ E, and let α ∈ E, α /∈ B. Then a = α p ∈ B so α is a root ofX p − a ∈ B[X ]. But, by Lemma 5.1.2, this polynomial is irreducible, so it isequal to mα(X), the minimum polynomial of α over B. Then this polynomialmα(X) = X p − a = X p − α p = (X − α)p has repeated roots, contradictingthe separability of E over B. ��Theorem 5.1.8. Let B be a field intermediate between F and E. If E is separa-ble extension of B and B is a separable extension of F, then E is a separableextension of F.

Proof. First, assume both (E/B) and (B/F) are finite. Since B is a separableextension of F, we have B = FBp, and since E is a separable extension of B,we also have E = BEp, both by Lemma 5.1.7.

Then

E = BEp = FBp(BEp)p ⊆ FEp,

so E = FEp and (E/F) = (E/B)(B/F) is finite, and so E is a separableextension of F by Lemma 5.1.3.

Now for the general case. Let α ∈ E, and let mα(X) ∈ B[X ] be the min-imum polynomial of α over B. Let B0 be the subfield of B obtained by ad-joining the coefficients of mα(X) to F. Then B0 is a finite extension of F andF ⊆ B0 ⊆ B with B a separable extension of F, by hypothesis, so B0 is aseparable extension of F by Lemma 5.1.6. Let E0 = B0(α). Then E0 is a finiteextension of B0. Since E is a separable extension of B, by hypothesis, mα(X)

is a separable polynomial, so, by Corollary 5.1.5, E0 is a separable extensionof B0. Then, by the finite case, E0 is a separable extension of F. Since α isarbitrary, E is separable. ��Theorem 5.1.9. A finite extension E of F is separable if and only if E is ob-tained from F by adjoining root(s) of a separable polynomial f (X) ∈ F[X ].Proof. E is obtained from F by successively adjoining finitely many elements.If one of them is not separable, then certainly E is not separable. If each ofthem is a root of a separable polynomial, then the theorem follows immediatelyfrom Corollary 5.1.5 and Theorem 5.1.8. ��Lemma 5.1.10. Let E be an extension of F. Then

Fs = {α ∈ E | α is separable over F}is a field intermediate between F and E.


Proof. Let α, β ∈ Fs . Then α is a simple root of mα(X) ∈ F[X ] ⊆ F(β)[X ],so F(β)(α) = F(α, β) is separable over F(β). Also, F(β) is separable over F.Hence, by Theorem 5.1.8, F(α, β) is separable over F. Thus α + β ∈ Fs andαβ ∈ Fs . Also, if α �= 0 then 1/α ∈ F(α) so 1/α ∈ Fs . Thus Fs is a field. ��Definition 5.1.11. The field Fs of Lemma 5.1.10 is called the separableclosure of F in E. �

We now turn our attention to inseparable extensions.

Definition 5.1.12. Let E be an extension of F. An element α ∈ E is purelyinseparable over F if mα(X) = (X − α)m ∈ E[X ] for some m. The integer mis called the degree of inseparability of α. �Remark 5.1.13. If α ∈ E is both separable and purely inseparable over F, then(m = 1 and) α ∈ F. �Lemma 5.1.14. Let α ∈ E be purely inseparable over F with degree of insep-arability m. Then m = pr for some r.

Proof. Write m = kpr with (k, p) = 1. Then, in E[X ],mα(X) = (X − α)kpr = ((X − α)pr

)k = (X pr − α pr)k

= (X pr)k − kα pr

(X pr)k−1 + · · · ∈ F[X ]

by the Binomial Theorem. Thus kα pr ∈ F and so α pr ∈ F. Thus f (X) =X pr − α pr ∈ F[X ] and f (X) divides mα(X) in E[X ] and hence in F[X ],by Lemma 2.2.7. Since mα(X) is irreducible in F[X ], mα(X) = f (X) andk = 1. ��Corollary 5.1.15. If α is purely inseparable over F, then (F(α)/F) = m = pr

for some r.

Proof. (F(α)/F) = deg mα(X). ��Definition 5.1.16. An extension E of F is purely inseparable if every α ∈ E isinseparable over F. �Lemma 5.1.17. Let E be an extension of F. If α ∈ E is purely inseparable,then F(α) is a purely inseparable extension of F.

Proof. First observe that mα(X) ∈ F[X ] factors as mα(X) = (X − α)prin

E[X ], by Lemma 5.1.14, so (X −α)pr = X pr −α pr ∈ F[X ] and a = α pr ∈ F.


Now let β ∈ F(α). If β ∈ F, there is nothing to prove (as then mβ(X) =X − β) so assume β /∈ F. Then β = ∑m−1

i=0 biαi with bi ∈ F (and m = pr ), so

β pr =( m−1∑

i=0

biαi)pr

=m−1∑i=0

bpr

i (α pr)i =

m−1∑i=0

bpr

i ai ∈ F,

so F = F(β pr) and hence F(β pr

) ⊂ F(β). Then, by Lemma 5.1.4, β is insep-arable over F. (If β were separable over F, we would have F(β) = F(β p) =F((β p)p) = · · · = F(β pr

), a contradiction.) ��Lemma 5.1.18. Let E be a purely inseparable extension of F. Then every α ∈E is purely inseparable over F.

Proof. Let α ∈ E. If α ∈ F, there is nothing to prove, so let α /∈ F. Considerits minimum polynomial mα(X) and let pe be the highest power of p dividingall the exponents of the nonzero terms in mα(X). Then mα(X) = f (X pe

)

for some polynomial f (X), and f (X) is irreducible as mα(X) is. Note thatf ′(X) �= 0 as some term in f (X) has positive degree not divisible by p, so,by Proposition 3.2.2, f (X) is separable. Now α pe

is a root of f (X), so α peis

a separable element of E. Hence a = α pe ∈ F, and α satisfies the polynomialX pe − a. Let f be the smallest integer with a′ = α p f ∈ F. Then a′ /∈ Fp,by Lemma 5.1.1, so, by Lemma 5.1.2, g(X) = X p f − a′ is irreducible, andg(α) = 0. Hence mα(X) = g(X) = X p f − a′ = (X − α)p f

and α is purelyinseparable (and f = e). ��Proposition 5.1.19. Let E be a purely inseparable extension of F and let Bbe any field intermediate between F and E. Then B is a purely inseparableextension of F and E is a purely inseparable extension of B.

Proof. Clearly B is a purely inseparable extension of F.Let α ∈ F. Then α has minimum polynomial mα(X) ∈ F[X ] which factors

as (X − α)m ∈ E[X ]. Let mα(X) ∈ B[X ] be the minimum polynomial of α

over B. Then mα(X) divides mα(X) in E[X ], so mα(X) = (X − α)m ′for

some m ′, and α is purely inseparable over B. Since α is arbitrary, E is purelyinseparable over B. ��Corollary 5.1.20. Let E be a finite, purely inseparable extension of F. Then(E/F) is a power of p.

Proof. Let {α1, . . . , αk} be a basis for E over F, so E = F(α1, . . . , αk). Weprove the corollary by induction on k.

If k = 1, then E = F(α1) and the result is immediate from Corollary5.1.15.


Assume the result is true for k − 1, and let B = F(α1, . . . , αk−1). Since Eis a purely inseparable extension of F, E is a purely inseparable extension ofB and B is a purely inseparable extension of F, by Proposition 5.1.19. Since(E/F) = (E/B)(B/F), the result follows. ��Lemma 5.1.21. Let E be an algebraic extension of F and let Fs be the sepa-rable closure of F in E. Then E is a purely inseparable extension of Fs .

Proof. Suppose α ∈ E is separable over Fs . Then Fs(α) is a separable exten-sion of Fs and Fs is a separable extension of F so, by Theorem 5.1.8, Fs(α) isa separable extension of F, and then α ∈ Fs by the definition of Fs . ��Corollary 5.1.22. Let E be an algebraic extension of F. Then there is a uniquefield B intermediate between F and E with B a separable extension of F and Ea purely inseparable extension of B.

Proof. Take B = Fs . ��Note that Corollary 5.1.22 says that any algebraic extension of F can be

obtained by first taking a separable extension, and then taking a purely insep-arable extension.

Definition 5.1.23. Let E be an algebraic extension of F and let Fs be the sep-arable closure of F in E. Then (Fs/F) is the separable degree of E over F and(E/Fs) is the inseparable degree of E over F. �Remark 5.1.24. (1) Of course, (E/F) = (E/Fs)(Fs/F).

(2) E is a separable extension of F if and only if E = Fs , or, equivalently,if and only if (Fs/F) = (E/F).

(3) E is a purely inseparable extension of F if and only if Fs = F, or,equivalently, if and only if (Fs/F) = 1.

(4) By Corollary 5.1.20, (E/Fs) is a power of p.(5) If (E/F) is relatively prime to p, then E is a separable extension of F (as

in this case (E/Fs), which is a power of p, divides (E/F), which is relativelyprime to p, so we must have (E/Fs) = 1 and hence E = Fs). �Remark 5.1.25. Observe that every purely inseparable extension of F is nor-mal. (If α ∈ E is purely inseparable, then mα(X) = (X − α)m ∈ E[X ] is aproduct of linear factors.) �Theorem 5.1.26. Let B be a purely inseparable extension of F and let E be apurely inseparable extension of B. Then E is a purely inseparable extensionof F.


Proof. Let α ∈ E and let mα(X) ∈ B[X ] be the minimum polynomial of α

over B. Let B0 be the extension of F obtained by adjoining the coefficientsof mα(X) to F. Then F ⊆ B0 ⊆ B so B0 is purely inseparable over F byProposition 5.1.19, and B0 is certainly a finite extension of F. We now useTheorem 5.2.1 below to conclude that B0 is the splitting field of a polynomialf (X) ∈ F[X ]. Let {β1, . . . , βk} be the roots of f (X), so B0 = F(β1, . . . , βk).

Note also that mα(X) = (X − α)m ∈ E[X ], so α is purely inseparableover B0.

Let E0 be a splitting field of mα(X) ∈ F[X ] with F(α) ⊆ E0. Let α′ ∈ E0

be any root of mα(X). Since mα(X) is irreducible, by Lemma 2.6.1 there isa σ : F(α) → F(α′) with σ | F = id and σ(α) = α′. Since B0 is the split-ting field of f (X), by Lemma 2.6.3 σ extends to σ : F(α)(β1, . . . , βk) →F(α′, β1, . . . , βk). Now F(α)(β1, . . . , βk) = F(β1, . . . , βk)(α) = B0(α). Sim-ilarly, F(α′, β1, . . . , βk) = B0(α

′). Now σ(βi ) = βi as σ must take a rootof mβi (X) to a root of mβi (X), and βi is the only root of mβi (X). Henceσ | B0 = id. Thus σ : B0(α) → B0(α

′) with σ | B0 = id, σ(α) = α′.Then 0 = σ(0) = σ(mα(α)) = mα(σ (α)) = mα(α′). But α is purely insepa-rable over B0 so the only root of mα(X) is α. Hence α′ = σ(α) = α, and so α

is the only root of mα(X) and hence α is purely inseparable over F. Since α isarbitrary, E is purely inseparable over F. ��

We now extend the second part of Definition 5.1.12.

Definition 5.1.27. Let E be an algebraic extension of F and let α ∈ E. Thedegree of inseparability i(α) is the multiplicity of α as a root of mα(X). �Lemma 5.1.28. (1) i(α) is independent of E.

(2) If α and α′ are two roots of the irreducible polynomial f (X) ∈ F[X ]in E, then i(α) = i(α′).

Proof. (1) Let E′ ⊇ E, i = i(α) in E, i ′ = i(α) in E′. In E[X ], mα(X) =(X −α)i g(X) with g(X) relatively prime to X −α, and then g(X) is relativelyprime to X − α in E′[X ] as well (Lemma 2.2.7), so i ′ = i .

(2) By Lemma 2.6.1, there is an isomorphism σ : F(α) → F(α′) withσ | F = id and σ(α) = α′. Then σ( f (X)) = f (X), and also if we writef (X) = (X − α)i g(X) ∈ F(α)[X ] with g(α) �= 0, then we see that f (X) =(X − α′)i g(X) ∈ F(α′)[X ] with g(α′) �= 0. ��Lemma 5.1.29. Let E be an algebraic extension of F and let α ∈ E. Then i(α)

is a power of p. More precisely, i(α) is the highest power of p dividing theexponent of every nonzero term in mα(X).

5.2 Normal Extensions 151

Proof. Let pr be the highest power of p such that f (X) = g(X pr), g(X) ∈

F[X ]. Since mα(X) is irreducible, so is g(X), and g′(X) �= 0, so g(X) isseparable (Proposition 3.2.2) and hence α pr

is a simple root of g(X). Henceg(X) = (X −α pr

)h(X) with h(α pr) �= 0, i.e., with X −α pr

not dividing h(X).Thus, one the one hand, X pr −α pr = (X −α)pr

divides f (X) and, on the otherhand, this is the highest power of X − α dividing f (X). ��Corollary 5.1.30. Let E be an algebraic extension of F and let α ∈ E. Ifmα(X) splits in E[X ], then

mα(X) = (X − α1)pr · · · (X − αk)

pr

with αi = α for some i , pr = i(α), and kpr = deg mα(X). ��

5.2 Normal Extensions

In this section we investigate normal extensions.

Theorem 5.2.1. E is a finite normal extension of F if and only if E is the split-ting field of a polynomial f (X) ∈ F[X ].Proof. If E is normal and {ε1, . . . , εn} is a basis for E over F, thenmεi (X) ∈ F[X ] splits into a product of linear factors for each i , so f (X) =mεi (X) · · · mεn (X) ∈ F[X ] splits into a product of linear factors, and then Eis the splitting field of f (X), as any field in which f (X) splits must contain{ε1, . . . , εn}.

Suppose now that E is the splitting field of f (X) ∈ F[X ] and let α1, . . . , αk

be the roots of f (X) in E, so E = F(α1, . . . , αk). Let g(X) ∈ F[X ] have a rootβ ∈ E. We need to show g(X) splits into a product of linear factors in E[X ].Clearly it suffices to consider the case that g(X) is irreducible.

Let E ⊇ E be a splitting field of the irreducible polynomial g(X) and letβ ∈ E be a root of g(X).

Then, by Lemma 2.6.1, there is a unique isomorphism σ : F(β) → F(β)

with σ(β) = β and σ | F = id.Now E is a splitting field for f (X) over F(β) and E(β) = F(α1, . . . , αk,

β) is a splitting field for f (X) over F(β) so by Lemma 2.6.3 there is an iso-morphism τ : E = E(β) → E(β) extending σ , and hence with τ(β) = β.

Now τ extends σ , which is the identity on F, so τ is itself the identity onF. Thus τ permutes {α1, . . . , αk}, as it must take any root of f (X) to a root off (X). Since β ∈ E = F(α1, . . . , αk), β = h(α1, . . . , αk) for some polynomialh(X1, . . . , Xk) ∈ F[X1, . . . , Xk]. Then β = τ(β) = τ(h(α1, . . . , αk)) =h(τ (α1), . . . , τ (αk)) so β ∈ E. ��


Corollary 5.2.2. Let E be a normal extension of F and let σ : B → B′ beany isomorphism of fields B and B′ intermediate between E and F such thatσ | F = id. Then σ extends to an automorphism σ : E → E.

Proof. First, suppose that E is a finite extension of F. Then E is the splittingfield of a polynomial f (X) ∈ F[X ]. We may regard E as the splitting fieldof f (X) ∈ B[X ] and note that σ( f (X)) = f (X). Then the corollary followsimmediately from Lemma 2.6.3.

The general case then follows by a Zorn’s Lemma argument. (See the proofof Lemma 5.4.1 below.) ��Definition 5.2.3. Let E be an algebraic extension of F. An extension D of E isa normal closure of E if D is a normal extension of F but no field intermediatebetween D and E is a normal extension of F. �Lemma 5.2.4. Every algebraic extension E of F has a normal closure D, andany two normal closures of E are isomorphic.

Proof. First, suppose E is a finite extension of F. Let E = F(α1, . . . , αk)

and let f (X) = mα1(X) · · · mαk (X). Let D ⊇ E be a splitting field of f (X).Then D is a normal extension of F by Theorem 5.2.1, and f (X) does notsplit in any proper subfield of D containing E, so D is a normal closure ofE. Furthermore, D is unique up to isomorphism as any two splitting fields off (X) are isomorphic (Lemma 2.6.4).

The general case then follows by a Zorn’s Lemma argument. ��Corollary 5.2.5. Let E be a separable extension of F. Then the normal closureD of E is a Galois extension of F.

Proof. First, suppose E is a finite extension of F. Then, following the proof ofLemma 5.2.4, D is a splitting field of the separable polynomial f (X) and so,by Theorem 2.7.14, D is a Galois extension of F.

The general case then follows from Corollary 5.4.3 below. ��This corollary justifies the following definition.

Definition 5.2.6. Let E be a separable extension of F. A normal closure D ofE is called a Galois closure of E. �Remark 5.2.7. (1) Note that if E is normal over F and B is an intermediate fieldbetween F and E, then E is normal over B, but B need not be normal over F.E is normal over B as for any α ∈ E, mα(X) splits into linear factors in E, andhence mα(X), the minimum polynomial of α over B, which divides mα(X),

5.2 Normal Extensions 153

also splits into linear factors in E. On the other hand, if E = Q(ω,3√

2) is thesplitting field of X3 − 2 ∈ Q[X ], and B = Q(

3√

2), F = Q, then E is normalover B but B is not normal over F.

(2) If B is a normal extension of F and E is a normal extension of B, thenE need not be a normal extension of F. Taking F = Q, B = Q(

√2), and

E = Q(4√

2) provides an example of this. �Now we turn to inseparable extensions. We first recall from Remark 5.1.25

that a purely inseparable extension is automatically normal.

Lemma 5.2.8. Let E be a normal extension of F. Then E is separable ifand only if Fix(Gal(E/F)) = F, and E is purely inseparable if and only ifFix(Gal(E/F)) = E, or, equivalently, if and only if Gal(E/F) = {id}.Proof. The first claim is immediate from the fact that E is a Galois extensionof F if and only if E is normal and separable (Theorem 2.7.14 in case (E/F)

is finite and Theorem 5.4.2 below in case (E/F) is infinite).For the second claim, first suppose that (E/F) is finite. Suppose next that E

is a purely inseparable extension of F. Let α ∈ E. Then mα(X) = (X −α)m inE[X ] for some m. Now for any σ ∈ Gal(E/F), σ(α) is also a root of mα(X).But the only root of mα(X) is α, so σ(α) = α. Since α is arbitrary, σ = id.

The general case follows by a Zorn’s lemma argument.

Conversely, let E be normal and finite over F, and suppose that E is notpurely inseparable over F. Then there is an α ∈ E such that mα(X) has (atleast) two distinct roots α and α′. By Lemma 2.6.1 there is a σ : F(α) →F(α′) with σ(α) = α′ and, by Corollary 5.2.2, σ extends to σ : E → E withσ (α) = α′ and σ | F = id, so σ ∈ Gal(E/F) and σ �= id. ��

We have seen in Section 5.1 that any algebraic extension may be obtainedby first taking a separable extension and then taking a purely inseparable ex-tension. In the case of a normal extension, the order may be reversed.

Proposition 5.2.9. Let E be a normal extension of F. If Fi = Fix(Gal(E/F))

then Fi is a purely inseparable extension of F and E is a separable extensionof Fi . Furthermore, E is a Galois extension of Fi and Gal(E/Fi ) = Gal(E/F).

Proof. By definition, if G is a group of automorphisms of E, then E is a Galoisextension of Fix(G). Apply that here with G = Gal(E/F). Then E is a Galoisextension of Fi and hence E is a separable extension of F (Theorem 2.7.14 incase (E/Fi ) is finite and Theorem 5.4.2 below in case (E/Fi ) is infinite).

Now we consider the extension Fi of F. If Fi is not purely inseparable then,as in the proof of Lemma 5.2.8, there is an element α ∈ Fi and an isomorphism


σ : F(α) → F(α′) with σ | F = id and σ(α) = α′ �= α. Then, by Corollary5.2.2, σ extends to σ : E → E with σ | F = id, i.e., to σ ∈ Gal(E/F) withσ (α) �= α, contradicting the definition of Fi . ��

In fact, in this situation we may say more. Recall that Fs was defined inDefinition 5.1.11 and characterized in Corollary 5.1.22.

Theorem 5.2.10. Let E be a normal extension of F. Then Fs and Fi are disjointextensions of F and E = FsFi . Furthermore, Fs is a Galois extension of F andGal(Fs/F) = Gal(E/Fi ). Also, (E/Fi ) = (Fs/F) and (E/Fs) = (Fi/F).

Proof. Let α ∈ Fs ∩ Fi . Then α is both separable and purely inseparable overF, so, by Remark 5.1.13, α ∈ F. Thus Fs and Fi are disjoint extensions of F.

Let α ∈ Fs . Then α has separable minimum polynomial mα(X). SinceE is a normal extension of F, mα(X) splits into a product of linear factors(X −α1) · · · (X −αt ) in E[X ]. But each αi is a root of a separable polynomial,namely mα(X), so αi ∈ Fs for each i , and hence mα(X) splits into a product oflinear factors (X −α1) · · · (X −αt ) in Fs[X ]. Thus Fs is a normal and separableextension of F, so is a Galois extension of F (by Theorem 2.7.14 in case E isa finite extension of F, and by Theorem 5.4.2 below in general).

Suppose that E is a finite extension of F. By Theorem 3.4.6, Gal(E/Fi ) isisomorphic to a subgroup of Gal(Fs/F). But let σ ∈ Gal(Fs/F). Now E is apurely inseparable extension of Fs , so it is a normal extension of F (Remark5.1.15) and hence it is the splitting field of some polynomial f (X) ∈ F[X ]by Theorem 5.2.1. Hence, by Lemma 2.6.3, σ extends to an automorphismof E, i.e., to an element of Gal(E/F) = Gal(E/Fi ), so these two groups areisomorphic. But then also

(E/F) = (E/Fi )(Fi/F) = (E/Fs)(Fs/F)

giving the claimed equalities between the degrees of the extensions. Further-more, since Fs is a Galois extension of F, we have from Theorem 2.4.5 that

(FsFi/F) = (Fs/F)(Fi/F) = (E/F),

so FsFi = E.

Now for the general case. Let α ∈ E and let E0 ⊆ E be the splittingfield of mα(X). Then E0 is a finite extension of F, so we may apply the aboveargument to E0, Fs ∩ E0, and Fi ∩ E0. Then we may apply Zorn’s Lemmato obtain the conclusion of the theorem for E. (The final point to see is thatwe have equalities among the degrees even if they are not all finite, whenwe cannot simply use the argument above. But (E/Fs) ≤ (Fi/F) by Lemma

5.3 The Algebraic Closure 155

2.3.4. If we did not have equality, then there would be some nontrivial linearcombination of basis elements of E over Fs that would be zero, and this linearcombination would involve only finitely many basis elements and hence wouldhave coefficients in some finite extension of F, contradicting equality of thedegrees in the finite case.) ��

5.3 The Algebraic Closure

In this section we show that every field F has an algebraic closure F that isunique up to isomorphism.

Definition 5.3.1. A field E is algebraically closed if the only algebraic exten-sion of E is E itself. �Lemma 5.3.2. The following are equivalent:

(1) E is algebraically closed.(2) Every nonconstant polynomial f (X) ∈ E[X ] is a product of linear

factors in E[X ].(3) Every nonconstant polynomial f (X) ∈ E[X ] has a root in E.

Proof. (1) implies (2): Let E be algebraically closed. Let f (X) ∈ E[X ] beirreducible. Then f (X) has a root α in some algebraic extension E(α) of E.But E(α) = E by assumption, so X −α divides f (X) in E and hence f (X) =X − α. Now let g(X) ∈ E[X ] be arbitrary. Then g(X) splits into a product ofirreducible and hence linear factors in E, yielding (2).

(2) implies (1): Let α be an element of some extension of E with α alge-braic over E. Then mα(X) = E[X ] splits into a product of linear factors in E,one of which must be X − α. Hence α ∈ E, yielding (1).

(2) implies (3): Let f (X) ∈ E[X ]. Then f (X) = (X − α1) · · · (X − αk)

for some α1, . . . , αk ∈ E, so f (α1) = 0, yielding (3).(3) implies (2): By induction on k = deg f (X). For k = 1 this is trivial.

Assume it is true for k − 1 and let f (X) have degree k. Then f (X) has a rootα1 so f (X) = (X − α1)g(X) with deg g(X) = k − 1. By the inductive hy-pothesis g(X) = (X −α2) · · · (X −αk) for some α2, . . . , αk , and then f (X) =(X − α1)(X − α2) · · · (X − αk), yielding (2). ��Definition 5.3.3. Let F be a field. An algebraic closure F of F is an algebraicextension of F that is algebraically closed. �Theorem 5.3.4. Let F be an arbitrary field. Then F has an algebraic closureF and F is unique up to isomorphism.


Proof. Both existence and uniqueness are Zorn’s Lemma arguments.The idea of the proof is clear. We wish to take the set of all algebraic ex-

tensions of F, order this set by inclusion, use Zorn’s Lemma to conclude thisset has a maximal element, and show this maximal element is an algebraicclosure of F. But, in carefully considering this argument, set-theoretic difficul-ties appear, as, a priori, it is not clear what inclusion means. (We can say, forexample, that Q(

√2) ⊂ Q(

√2,

√3) as we can regard these both as subfields

of C. But this is because we already know that C exists. In the abstract, whatwe mean by this inclusion is the following: Let F1 be the field obtained fromQ by adjoining an element α with α2 = 2 and let F2 be the field obtained fromQ by adjoining elements β1 and β2 with β2

1 = 2 and β22 = 3. Then there is

a map σ : F1 → F2 given by σ(α) = β1 and using this map we may identifyF1 with a subfield of F2. But this is not quite the same thing as saying that F1

is a subset of F2.) The way out of this dilemma is to assume that all fields weneed to consider are contained in some universal set. (Note we cannot assumethey are all contained in some universal field as that would be presuming theexistence of the algebraic closure, which is what we are trying to prove.)

Thus we proceed as follows in order to show the existence of F.Given a set U , and a field E, we shall write E ⊆ U if E is a subset of U .

Then the field operations (addition, multiplication, inversion) are defined onthis subset of U . For two such fields E1 and E2, we shall write E1 ⊆ E2 ifE1 ⊆ E2 as subsets of U and the restrictions of the field operations on E2 toE1 agree with the field operations on E1.

We now set U0 = 2F, the set of subsets of F. If F is infinite we let U = U0.If F is finite we let U = 2U0 . Regard F ⊂ U by identifying a with {a}, a ∈ F,if F is infinite, or with {{a}}, if F is finite.

As is well known, for any set A, the cardinality of 2A is greater than thecardinality of A, so the cardinality of F is less than the cardinality of U . Fur-thermore, by construction, U is uncountable.

Let

S = {algebraic extensions E of F with E ⊆ U}and order S by inclusion. Then every chain has a maximal element: If E1 ⊆E2 ⊆ · · · , then E = ∪Ei is algebraic. (Any α ∈ E is an element of some Ei

and hence is algebraic over F.) Hence, by Zorn’s Lemma, S has a maximalelement F.

Claim: F is algebraically closed.

Proof of claim: Let F ⊆ F with F an algebraic extension of F. Then, byTheorem 2.4.12, F is an algebraic extension of F. To complete the existence


part of the proof we must show that it is possible to embed F in U . Assumingthat, by the maximality of F we must have that this image is just F, and thenF = F, so we see that F is algebraically closed.

Claim: Let E be an algebraic extension of F. If F is finite, E is finite orcountably infinite. If F is infinite, E has the same cardinality as F.

Assuming this claim, F ⊂ U is a set of smaller cardinality than that of Uand F ⊂ F with F also a set of smaller cardinality than U , so we may choose anarbitrary embedding of F − F in U as a set and then define the field operationson this set to agree with those on F.

Proof of claim: We consider the set P of all monic polynomials in F[X ].Then P = ⋃∞

n=1 Pn , where Pn is the set of all monic polynomials of degreen. Now there is a bijection between Fn and Pn given by (a0, . . . , an−1) ←→a0 + a1 X + · · · + an−1 Xn−1 + Xn . If F is finite, so is Fn for each n, and henceP is countable union of finite sets and so is countable. If F is infinite, then Fn

has the same cardinality as F for each n, and then P is a countable union ofsuch sets, so also has the same cardinality as F.

Now if E is an algebraic extension of F, we have a map ρ : E → P givenby ρ(α) = mα(X). Then ρ is a finite-one mapping as if ρ(α) = f (X), thenf (α) = 0, and any polynomial only has finitely many roots. But, in general, ifρ : A → B is a finite-to-one mapping of the set A onto the infinite set B, thenthe cardinality of A is equal to the cardinality of B.

This completes the proof of the existence of F. Now we must show that Fis unique up to isomorphism.

Thus let F1 and F2 be two algebraic closures of F. Let

S = {(E, σ ) | E ⊆ F1, σ : E → F2}where σ is a map of fields.

Order S by (E1, σ1) ≤ (E2, σ2) if E1 ⊆ E2 and σ2 | E1 = σ1. Again byZorn’s Lemma every chain has a maximal element so S has a maximal element(F1, σ ).

Claim: F1 = F1.

Proof of claim: Suppose not, and let α ∈ F1, α1 /∈ F1. Let σ (F1) = F2 ⊆F2. Since α1 is algebraic over F, it is algebraic over F1, so let mα1(X) ∈ F1(X)

be its minimum polynomial over F1. Since F2 is algebraically closed, thereis a root α2 of σ(mα1(X)) in F2. Then, by Lemma 2.6.1, σ extends to a mapτ : F1(α1) → F2(α2) ⊆ F2 (with τ (α1) = α2), contradicting the maximalityof F1. Thus F1 = F1 and we have a map σ : F1 → F2.

As σ is a map of fields, it is injective. To complete the proof we must showit is surjective. Let β2 ∈ F2. Then β2 is algebraic over F, so it is algebraic over


σ(F1). But σ(F1) is algebraically closed so β2 ∈ σ(F1), i.e., β2 = σ(β1) forsome β1 ∈ F1, as required. ��

We now wish to examine some concrete algebraically closed fields. Butfirst we shall prove a lemma that enables us to weaken the hypotheses inLemma 5.3.2.

Lemma 5.3.5. Let E be an algebraic extension of F. Then E is algebraicallyclosed if and only if every polynomial f (X) ∈ F[X ] is a product of linearfactors in E[X ]. If E is a normal extension of F, then E is algebraically closedif and only if every polynomial f (X) ∈ F[X ] has a root in E.

Proof. First, let E be an algebraic extension of F. Then the “only if” part isclear. For the “if” part, let α be algebraic over E. Then E(α) is an algebraicextension of E (Corollary 2.4.9) and E is algebraic over F (by hypothesis),so E(α) is algebraic over F (Theorem 2.4.12). Hence α is a root of mα(X) ∈F[X ]. But mα(X) = (X − α1) · · · (X − αn) in E[X ], so α = αi ∈ E for somei . Now if E is a normal extension of F, then by definition, any polynomial inF[X ] with a root in E splits into a product of linear factors in E[X ], so we areback in the previous case. ��Theorem 5.3.6. The field of rational numbers Q has an algebraic closure Qthat is a subfield of the field of complex numbers C.

Proof. Let f (X) ∈ Q[X ] be an arbitrary irreducible polynomial and let Ebe a splitting field of f (X). Let G = Gal(E/Q), and let H be the 2-Sylowsubgroup of G. Set B = Fix(H). Then E is a Galois extension of B withGalois group H . Also, d = [G : H ] = (B/Q) is odd. By Corollary 3.5.3,there is an element α of B with B = Q(α). Then mα(X) ∈ Q[X ] is a monicirreducible polynomial of odd degree d.

By elementary calculus, mα(X) has a real root x0. (Proof: limx→−∞ mα(x)

= −∞, so mα(x−) < 0 for some x− ∈ R, and limx→∞ mα(x) = ∞, somα(x+) > 0 for some x+ ∈ R. Then, by the Intermediate Value Theorem,mα(x0) = 0 for some x0 ∈ R.)

Let B′ = B(x0) ⊆ R. Then f (X) has a splitting field E′ ⊇ B′. Let H ′ =Gal(E′/B′). By Corollary 3.4.6 (the Theorem on Natural Irrationalities), H ′ isisomorphic to a subgroup of H , so H ′ is also a 2-group. Thus, by CorollaryA.2.3, there is a sequence H ′ = H ′

0 ⊃ H ′1 ⊃ · · · ⊃ H ′

n = {1} with [Hi−1 :Hi ] = 2, and, setting E′

i = Fix(H ′i ), B′ = E′

0 ⊂ E′1 ⊂ . . . E′

n = E′ with(E′

i/E′i−1) = 2. In other words, for each i , E′

i is a quadratic extension of E′i−1,

i.e., E′i = E′

i−1(√

αi ) for some αi ∈ E′i−1. But if F ⊆ C and D = F(

√z)

for some z ∈ F, then D is isomorphic to a subfield of C. (If z = reiθ , D =


F(√

reiθ/2).) Thus, proceeding inductively, we see that f (X) has a splittingfield that is a subfield of C.

Now let { fi (X)} be the set of all monic irreducible polynomials in Q[X ]and let Ei ⊆ C be the splitting field of fi (X). Let Q be the composite of {Ei }.Then Q is algebraic over F and every f (X) ∈ Q[X ] splits in Q, so, by Lemma5.3.5, Q is an algebraic closure of Q. ��

Actually, a similar argument enables us to prove a stronger result.

Theorem 5.3.7 (Fundamental Theorem of Algebra). The field of complexnumbers C is algebraically closed.

Proof. First, observe that it suffices to prove that every polynomial f (X) ∈R[X ] has a root in C. On the one hand, this follows directly from Lemma 5.3.5,but on the other hand, we have the following direct argument in this case: Letg(X) ∈ C[X ] be an arbitrary polynomial, and set f (X) = g(X)g(X). Thenf (X) ∈ R[X ]. If f (X) has a root γ , then g(γ ) = 0, in which case γ is a rootof g(X), or g(γ ) = 0, in which case g(γ ) = 0 and γ is a root of g(X).

Thus let f (X) ∈ R[X ] be an arbitrary irreducible polynomial and let Ebe a splitting field of f (X). Let G = Gal(E/R), and let H be the 2-Sylowsubgroup of G. Set B = Fix(H). Then E is a Galois extension of B with Galoisgroup H . Also, d = [G : H ] = (B/Q) is odd. By Corollary 3.5.3, there isan element α of B with B = R(α). Then mα(X) ∈ R[X ] is an irreduciblepolynomial of odd degree d. But, by the same elementary calculus argumentas in the proof of Theorem 5.3.6, mα(X) has a real root x0. Thus (X − x0)

divides mα(X). Since mα(X) is irreducible, we must have mα(X) = (X − x0),and so d = 1 and B = R.

Then E is a splitting field of f (X) with (E/R) = 2n for some n, and,arguing as in the proof of Theorem 5.3.6, we see that E is isomorphic to asubfield of C, so f (X) has a root in C, and we are done. ��Remark 5.3.8. (1) Q has an algebraic closure Q ⊆ C but Q �= C as Q iscountable, being an algebraic extension of Q, but C is uncountable.

(2) Theorem 5.3.7 implies Theorem 5.3.6, for, once we know C is alge-braically closed, it is easy to see that {z ∈ C | z is algebraic over Q} is analgebraic closure of Q. This field is known as the field of algebraic numbers.

(3) The Fundamental Theorem of Algebra cannot have a purely algebraicproof, as the definitions of R and C are not purely algebraic. We have givena proof that uses only a minimum of analysis (the argument from elementarycalculus given in the proof of Theorem 5.3.6). But this important theorem hasmany proofs. Here is one well-known proof that uses results from complexanalysis:


Let f (X) ∈ C[X ] be a nonconstant polynomial, f (X) = ∑ni=0 ai Xi . Then

for any z ∈ C, | f (z)| ≥ |an||z|n − ∑n−1i=0 |ai ||z|i from which it easily follows

that there is an A such that |z| ≥ A implies | f (z)| ≥ 1. Now suppose f (X)

does not have a root in C. Consider g(X) = 1/ f (X). Then |g(z)| ≤ 1 for|z| ≥ A. On the other hand, {z | |z| ≤ A} is a compact set so there is aC with |g(z)| ≤ C for all Z with |z| ≤ A. Thus g(z) is a bounded analyticfunction and hence, by Liouville’s theorem, a constant. But that implies f (z)is a constant, which is absurd. �

5.4 Infinite Galois Extensions

In this section we will consider infinite algebraic extensions. Our main goalswill be to show that Theorem 2.7.14 and Theorem 2.8.8 (the FundamentalTheorem of Galois Theory) have appropriate generalizations to this case.

First, we must generalize Lemma 2.6.3.

Lemma 5.4.1. Let E be a normal extension of F and let σ : B → B′ be anyisomorphism with σ | F = id, where B and B′ are any fields intermediatebetween F and E. Then σ extends to an automorphism of E, i.e., to an elementof Gal(E/F).

Proof. This is an application of Zorn’s Lemma. Let

S = {(D, τ ) | B ⊆ D, τ : D → D, τ | B = σ }.Order S by (D1, τ1) ≤ (D2, τ2) if D1 ⊆ D2 and τ2 | D1 = τ1. Then every

chain {(Di , τi )} in S has a maximal element (D, τ ) given by

D =⋃

Di , τ (d) = τi (d) if d ∈ Di .

Hence by Zorn’s Lemma S has a maximal element (E0, τ0). We claimE0 = E. Suppose not and let α ∈ E, α /∈ E0. Consider its minimum polyno-mial mα(X) over F. Since mα(X) ∈ F[X ], τ0(mα(X)) = mα(X).

Now consider mα(X) as an element of E0[X ], and let E1 ⊆ E be thesplitting field of mα(X) over E0. Note E0 ⊂ E1, as α /∈ E0 but α ∈ E1. Then,by Lemma 2.6.3, there is an automorphism τ1 of E1 extending τ0, contradictingthe maximality of (E0, τ0). ��

Now we generalize Theorem 2.7.14.

5.4 Infinite Galois Extensions 161

Theorem 5.4.2. Let E be an algebraic extension of F. The following are equiv-alent:

(1) E is a Galois extension of F.(2) E is a normal and separable extension of F.(3) Every field B intermediate between F and E with B a finite extension of

F is contained in a field D intermediate between E and F that is the splittingfield of a separable polynomial in F[X ], or, equivalently, that is a finite Galoisextension of F.

Proof. Although this not the logically most economical proof, we shall firstshow that (1) and (2) are equivalent and then that (3) is equivalent to both ofthese.

First, we show that (1) implies (2). Let α ∈ E. Then α is algebraic over F,so has minimum polynomial mα(X). Let {α1, . . . , αk} be the roots of mα(X) inE, with α1 = α. For any σ ∈ Gal(E/F), mα(σ (α)) = σ(mα(α)) = σ(0) = 0so σ(α) ∈ {α1, . . . , αk}, and thus the action of Gal(E/F) is to permute{α1, . . . , αk}. Consider f (X) = ∏k

i=1(X − αi ). This polynomial is invari-ant under Gal(E/F) so f (X) ∈ F[X ], as E is a Galois extension of F. Ifg(X) ∈ F[X ] is any polynomial with g(α) = 0, then by the same logicg(αi ) = 0 for each i , so f (X) divides g(X) in E[X ] and hence in F[X ].Thus f (X) is irreducible and so f (X) = mα(X). Hence mα(X) is a separablepolynomial that splits in E.

Next, we show that (2) implies (1). Let α ∈ E, and consider mα(X) ∈F[X ]. Then mα(X) = ∏k

i=1(X −αi ) in E[X ] for some {α1, . . . , αk} with α1 =α. Thus if we let B = F(α1, . . . , αk), we have that mα(X) = ∏k

i=1(X − αi )

in B[X ]. Thus we see that B is a finite extension of F and B is the split-ting field of a separable polynomial, so B is a Galois extension of F. HenceFix(Gal(B/F)) = F. Now let α /∈ F. Then α /∈ Fix(Gal(B/F)), and so thereis an automorphism σ of B with σ |F = id and with σ(α) �= α. By Lemma5.4.1, σ extends to an element τ ∈ Gal(E/F), and τ(α) = σ(α) �= α, so α /∈Fix(Gal(E/F)). Hence, as α /∈ F is arbitrary, we have that Fix(Gal(E/F)) = F,i.e., E is a Galois extension of F.

Next, we show that (2) implies (3). Since B is finite, B = F(α1, . . . , αk) forsome finite set {α1, . . . , αk}. Then mαi (X), i = 1, . . . , k, are all separable, andall split in E. Let D be the splitting field of the polynomial mα1(X) · · · mαk (X).Then B ⊆ D ⊆ E and D is a finite extension of F that is the splitting field of aseparable polynomial.

Finally, we show that (3) implies (1). Let α ∈ E, α /∈ F. Let B = F(α).Then B is a finite extension of F, so B ⊆ D where D is the splitting field of aseparable polynomial f (X) ∈ F[X ]. (In fact, we may take f (X) = mα(X).)


Thus D is a Galois extension of F, so there is an automorphism σ ∈ Gal(D/F)

with σ(α) �= α. By Lemma 5.4.1, σ extends to an element τ ∈ Gal(E/F), andτ(α) = σ(α) �= α, so α /∈ Fix(Gal(E/F)). Hence, as α /∈ F is arbitrary, wehave that Fix(Gal(E/F)) = F, i.e., E is a Galois extension of F. ��Corollary 5.4.3. Let E be an algebraic extension of F. The following areequivalent:

(1) E is a Galois extension of F.

(2) E is the splitting field of a set of separable polynomials in F[X ].Proof. Suppose that E is a Galois extension of F. Then, by Theorem 5.4.2, Eis normal and separable. Thus for every α ∈ E, mα(X) ∈ F[X ] is a separa-ble polynomial that splits in E[X ]. Then E is the splitting field of the set ofseparable polynomials {mα(X) | α ∈ E}.

On the other hand, let E be the splitting field of { fi (X)}i∈I , a set of sep-arable polynomials in F[X ]. First, observe that any ε ∈ E is contained in thesplitting field of a finite subset of { fi (X)}i∈I .

Now let B be any field intermediate between F and E that is a finite exten-sion of F. Then B = F(ε1, . . . , εn) for some finite set of elements {ε1, . . . , εn}of E, so, by the above observation, B is contained in the splitting field D ⊆ Eof a finite subset { f1(X), . . . , fk(X)} of { fi (X)}i∈I . Then D is the splittingfield of the separable polynomial f (X) = f1(X) · · · fk(X) ∈ F[X ], so, byTheorem 5.4.2, E is a Galois extension of F. ��Corollary 5.4.4. Let E be a Galois extension of F and let B be any field inter-mediate between F and E. Then E is a Galois extension of B.

Proof. This follows from Theorem 5.4.2. We restate condition (3) of thistheorem in order to avoid confusion of notation: Every field A intermediatebetween F and E with A a finite extension of F is contained in a field D inter-mediate between E and F that is the splitting field of a separable polynomial.We want to show this remains true when F is replaced by B.

Let A be a finite extension of B, so that A = B(α1, . . . , αn). Let A′ =F(α1, . . . , αn). We claim that A′ is a finite extension of F. (Of course, if B isa finite extension of F, then A is a finite extension of F and A′ ⊆ A, so thisis clear, but we need to prove this also in case B is not a finite extension ofF.) Since A′ = F(α1) · · · F(αn), in order to prove this it suffices to prove thatF(αi ) is a finite extension of F, for each i . Thus, fix i and consider F(αi ). Letmαi (X) ∈ B[X ] be the minimum polynomial of αi regarded as an element ofB. Let Bi be the field obtained by adjoining the coefficients of mαi (X) to F.Then Bi (α) is a finite extension of Bi and Bi is a finite extension of F, so Bi (αi )


is a finite extension of F, and F(αi ) ⊆ Bi (αi ), so F(αi ) is a finite extension ofF, as claimed.

Then, arguing as in the proof of Corollary 5.4.3, A′ ⊆ D′ ⊆ E, whereD′ is an extension of F that is the splitting field of a separable polynomialf (X) ∈ F[X ]. But A = BA′ ⊆ BD′, and BD′ is an extension of B that is thesplitting field of the separable polynomial f (X) ∈ B[X ], as required. ��

In order to generalize Theorem 2.8.8, the FTGT, we must introduce atopology on the Galois group.

Definition 5.4.5. Let E be an algebraic extension of F, and let G = Gal(E/F).The Krull topology on E is defined as follows: The identity element of G hasa neighborhood basis {UB = Gal(E/B) | B is a finite extension of F}. Anelement σ ∈ G has neighborhood basis {σUB}. �Remark 5.4.6. We explicitly observe that τ ∈ UB if and only if τ | B = id, andτ ∈ σUB if and only if τ | B = σ | B. �Remark 5.4.7. If E is a finite extension of F, then we may let B = E in theabove definition and we see that every element σ of G has a neighborhoodconsisting of σ alone. In other words, in this case G has the discrete topology.All of the results in this section remain true in this case, but the topology on Gyields no additional information. �

Definition 5.4.5 is not complete until we show these sets define a topologyon G. In fact we show a bit more.

Proposition 5.4.8. The sets {σUB} define a topology on G. Under this topol-ogy, G is a topological group.

Proof. Let Ui = UBi , i = 1, 2, and let U3 = U1 ∩ U2. Let B3 = B1B2. ThenB3 is a finite extension of F, and, by Proposition 2.8.14, B3 = Fix(U1 ∩U2) =Fix(U3). Furthermore, it is easy to check that if σ /∈ U3, then σ | B3 is not theidentity. Hence U3 = Gal(E/B3), i.e., U3 = UB3 . This shows that {UB} forma neighborhood basis of the identity, and hence that {σUB} form a basis for atopology on G.

Note that if σ ∈ G, then σUBσ−1 = Uσ(B) and σ(B) is a finite extensionof F. Then UBσ = σ(σ−1UBσ) = σUσ−1(B) so

{σUB | σ ∈ G, B/F finite} = {UBσ | σ ∈ G, B/F, finite}so, in defining the topology, we may take left cosets, right cosets, or both leftand right cosets.


To show G is a topological group we must show multiplication and in-version are continuous. If m = G × G → G is multiplication, and στU is aneighborhood of στ , then m−1(στU ) ⊃ (σ (τUτ−1))×(τU ), a neighborhoodof (σ, τ ) in G × G, and if i : G → G is inversion, and σ−1U is a neighbor-hood of σ−1, then i−1(σ−1U ) = U−1σ−1 = Uσ−1, a neighborhood of σ−1

in G. ��Lemma 5.4.9. Let E be a Galois extension of F and let B be any field inter-mediate between F and E. Then the Krull topology in Gal(E/B) is the inducedtopology on Gal(E/B) as a subspace of Gal(E/F) with the Krull topology.

Proof. Recall that, by Corollary 5.4.4, E is a Galois extension of B. Let U bea neighborhood of the identity in Gal(E/B), so U = Gal(E/A) for some finiteextension A of B. Then A = B(α) for some α ∈ E, by Corollary 3.5.3. LetD = F(α), so A = BD, and let U = Gal(E/D). Then

U = Gal(E/A) = Gal(E/BD) = Gal(E/B) ∩ Gal(E/D) = Gal(E/B) ∩ U,

and, similarly, if U is a neighborhood of the identity in Gal(E/F), then U =Gal(E/D) with D a finite extension of F, so

U = Gal(E/B) ∩ U = Gal(E/B) ∩ Gal(E/D) = Gal(E/BD)

is a neighborhood of the identity in Gal(E/B), as BD is a finite extensionof B. ��

Now we generalize the Fundamental Theorem of Galois Theory (Theorem2.8.8) to infinite Galois extensions.

Theorem 5.4.10 (Fundamental Theorem of Infinite Galois Theory). Let Ebe a Galois extension of F and let G = Gal(E/F) have the Krull topology.

(1) There is a 1-1 correspondence between intermediate fields E ⊇ B ⊇ Fand closed subgroups {1} ⊆ GB ⊆ G given by

B = Fix(GB).

(2) B is a normal extension of F if and only if GB is a normal subgroup ofG. This is the case if and only if B is a Galois extension of F. In this case thereis an isomorphism of topological groups

Gal(B/F) ∼= G/GB.


Proof. (1) For each closed subgroup H of G, let

BH = Fix(H).

This gives a map

� : {closed subgroups of G} → {fields intermediate between F and E}.We show � is a 1–1 correspondence.In this case it is easiest to work with the inverse correspondence. Thus, for

each field B intermediate between F and E, we let

GB = �(B) = Gal(E/B).

First, we claim that GB is a closed subgroup of G.Let σ ∈ GB, the closure of GB. Let β ∈ B be arbitrary. Then F(β) is a

finite extension of F, so σ has a neighborhood σN where N = Gal(E/F(β)).Since every neighborhood of σ has a nonempty intersection with GB, we maychoose τ ∈ GB ∩ σ N . Thus τ = σν for some ν ∈ N , and so, in particular,τ(β) = σν(β). Since τ ∈ GB, τ(β) = β, and, by the definition of N , ν(β) =β. Then σ(β) = σ(ν(β)) = σν(β) = τ(β) = β. Since β ∈ B is arbitrary,we see that σ fixes B. Then, since σ ∈ GB is arbitrary, we see that GB fixesB. Now by definition GB consists of all automorphisms of E fixing B. HenceGB = GB and so GB is closed.

Thus we have a map

� : {fields B intermediate between F and E} → {closed subgroups of G}and this map is 1–1 since if GB1 = �(B1) = �(B2) = GB2 , then, by Corol-lary 5.4.4, E is a Galois extension of B1 and of B2, so B1 = Fix(GB1) =Fix(GB2) = B2.

It remains to show � is onto. Thus, let H be a closed subgroup of Gand let B = Fix(H). Then K = Gal(E/B) ⊇ H and we need only showthey are equal. Let σ ∈ K be arbitrary and consider an arbitrary basic openneighborhood σU of σ . We claim σU ∩ H �= ∅. Assuming this claim, we havethat σ is in the closure H of H . Hence, as σ ∈ K is arbitrary, K ⊆ H . But His closed, so H = H and hence K = H .

To see the claim, let U = Gal(E/A) with A a finite extension of F. ByCorollary 5.4.4, E is a Galois extension of B, and, by Lemma 2.3.4, AB isa finite extension of B, so, by Theorem 5.4.2, AB is contained in a field Dthat is the splitting field of a separable polynomial f (X) ∈ B[X ]. Thus everyelement of Gal(E/B) restricts to an element of Gal(E/D), as it must take any


root of f (X) to a root of f (X), and conversely any element of Gal(D/B)

extends to an element (not, in general, a unique element, but that is irrelevant)of Gal(E/B), by Lemma 5.4.1. In other words, restriction to D give a well-defined epimorphism R : K = Gal(E/B) → Gal(D/B). Let σ0 = R(σ ) ∈Gal(D/B).

Now B ⊆ AB ⊆ D and D is a finite Galois extension of B. Let H0 =R(H) ⊆ Gal(D/B). Then Fix(H0) = Fix(H) = B = Fix(Gal(D/B)), so bythe Fundamental Theorem of (finite) Galois Theory, H0 = Gal(D/B). Thusthere is an element τ ∈ H with R(τ ) = σ0 = R(σ ), i.e., with τ | B = σ | B,and we see, by Remark 5.4.6, that τ ∈ σUB, so σU ∩ H �= ∅, as claimed.

(2) We have proved much of this in proving (1). Just as in the finite case,for any σ ∈ Gal(E/F), σ(B) = Fix(σ GBσ−1), so if GB is a normal subgroupof G, then σ(B) = B for every σ ∈ G, while if GB is not a normal subgroupof G, then σ(B) �= B for some σ ∈ G, and conversely. If B is a normalextension of F, then for every β ∈ B, mβ(X) ∈ F[X ] splits in B. Now for anyσ ∈ Gal(E/F), β ′ = σ(β) is a root of σ(mβ(X)) = mβ(X), so β ′ ∈ B, and soσ(B) = B. On the other hand, if B is not a normal extension of F, then thereis some β ∈ B such that mβ(X) has a root β ′ ∈ E, β ′ /∈ B, and then thereis an element σ of Gal(E/F) with σ(β) = β ′, by Lemma 5.4.1, and henceσ(B) �= B.

Now if B is normal, then, just as before, R : Gal(E/F) → Gal(B/F), givenby R(σ ) = σ | B, is a well-defined epimorphism, whose kernel is clearlyGB = Gal(E/B). It remains only to show that the Krull topology on Gal(B/F)

is the quotient topology on G/GB.Let U0 be a basic neighborhood of the identity in Gal(B/F). Then U0 =

{σ ∈ Gal(B/F) | σ | A = id} for some finite extension A of F (where ofcourse A ⊆ B) and U0 is the image under the quotient map R of the basicneighborhood of the identity UA ⊆ Gal(E/F). On the other hand, if UA is abasic neighborhood of the identity in Gal(E/F), its image under the quotientmap R is U0 = {σ ∈ Gal(B/F) | σ | B ∩ A = id}, and B ∩ A ⊆ A is a finiteextension of F, so U0 is a basic neighborhood of the identity in Gal(B/F). ��Proposition 5.4.11. Let E be a Galois extension of F and let B be an interme-diate field between F and E. Let G = Gal(E/F). Then:

(1) (E/B) is finite if and only if |GB| is finite. In this case, (E/B) = |GB|.(2) (B/F) is finite if and only if [G : GB] is finite. In this case, (B/F) =

[G : GB].(3) GB is always a closed subgroup of G. GB is an open subgroup of G if

and only if [G : GB] is finite.

Proof. (1) This follows immediately from Theorem 2.8.5.


(2)Assume B/F is finite and let B ⊆ D with D/F finite Galois (The-orem 5.4.2). Then, by the FTGT, (B/F) = [Gal(D/F) : Gal(D/B)]. ButGal(E/F) → Gal(D/F) is an epimorphism so

[G : GB] = [Gal(E/F) : Gal(E/B)] = [Gal(D/F) : Gal(D/B)] = (B/F).

Assume B/F is infinite and let {βi } be an infinite set of elements of B withF ⊂ F(β1) ⊂ F(β1, β2) ⊂ · · · ⊂ B. By Lemma 5.4.1, for each i = 1, 2, . . .

there is an element σi ∈ Gal(E/F) with σi (β j ) = β j for j < i but σi (βi ) �= βi ,so we see that σi �= σ j for i �= j . Since two elements of Gal(E/F) are in thesame left coset of GB if and only if they agree on B, we see that [G : GB] isinfinite.

(3) The fact that GB is closed is part of Theorem 5.4.10 (1).Suppose [G : GB] is finite and let {σ1, . . . , σk} be a set of left coset

representatives of GB with σ1 = 1. Since GB is closed, the finite unionσ2GB ∪ · · · σk GB is closed and hence its complement σ1GB = GB is open.

Suppose GB is open. Then it contains a basic open set UA = Gal(E/A),with A a finite extension of F. Then B = Fix(GB) ⊆ Fix(UA) = A, so (B/F)

is finite and hence [G : GB] is finite. ��We now further investigate the Krull topology on G.

Lemma 5.4.12. Each basic open set σUB in G is also closed.

Proof. Let τ /∈ σUB. Then ∅ = τB ∩ σUB (as UB is a subgroup of G), i.e.,τUB ⊆ G − σUB. In other words, the complement of σUB contains a neigh-borhood of each of its points, so is an open set, and hence σUB is a closedset. ��

We now prove a lemma we will need in describing the topology ofGal(E/F), but one which is interesting in its own right.

Lemma 5.4.13. Let E be an algebraic extension of F and let σ : E → E be amap of fields with σ | F = id. Then σ is an automorphism of E (and henceσ ∈ Gal(E/F)).

Proof. Since σ(1) = 1, σ is an injection, so we need only show σ is a surjec-tion. Let α ∈ E be arbitrary. Set

S = {roots of mα(X) in E} = {β ∈ E | mα(β) = 0}.Then for any β ∈ S, mα(σ (β)) = σ((mα(β)) = σ(0) = 0 so σ(β) ∈ S.

Thus σ | S is an injective map from S to itself. Since S is finite (as anypolynomial has only finitely many roots), σ | S is surjective as well, so thereis a β0 ∈ S with σ(β0) = α, as required. ��


Remark 5.4.14. This lemma is false for arbitrary extensions. For example, letE = F(X), the field of rational functions over F in the variable X . Thenσ(X) = X2 induces σ : E → E, an injective, but not surjective, map offields. �Theorem 5.4.15. Let E be a Galois extension of F and let G = Gal(E/F) havethe Krull topology. Then G is Hausdorff, totally disconnected, and compact.

Proof. We shall prove this in several steps.(1) G is Hausdorff: Since G is a topological group, to show G is Hausdorff

it suffices to show that⋂

UB = {id}where the intersection is taken over all finite extensions B of F.

Let σ ∈ G, σ �= id. Then for some α ∈ E, σ(α) �= α. Let B = F(α). Thenσ /∈ UB, as required.

(2) G is totally disconnected: Let σ, τ ∈ G, σ �= τ . Then τ /∈ σUB forsome UB, and then G = σUB ∪ (G − σUB) is a union of two relatively opensets with σ (resp. τ ) in the first (resp. second). (σUB is open by definitionand G − σUB is open as in the proof of Lemma 5.4.12.) Thus σ and τ are indifferent components of G, so the only components of G are single points.

(3) G is compact: This will require considerably more work.For each α ∈ E, let Rα = {β ∈ E | mα(β) = 0}. Let

G =∏α∈E

Rα.

By identifying an element∏

α∈E(βα)α with f : E → E by f (α) = βα foreach α ∈ E, we may regard G as a set of functions from E to itself. (This isidentifying a function with its graph.) Note that Rα = {α} for α ∈ F, so everyf ∈ G satisfies f (α) = α for every α ∈ F. Note also that Rα is finite foreach α. We give E the discrete topology, and each Rα the induced topology asa subset of E, which is also the discrete topology. Then Rα is a finite discretespace, so is compact. We give G the product topology. Then, by Tychonoff’sTheorem, G is the product of compact sets and hence is compact. (Recall thatif X = ∏

Xi is a product of spaces, the product topology has a basis of opensets

∏Yi where each Yi is open and Yi = Xi for all but finitely many i . If

each Xi is discrete, then every subset of Xi is open, so this condition reducesto Yi = Xi for all but finitely many i .)

Since every σ ∈ Gal(E/F) satisfies mα(σ (α)) = 0 for every α ∈ E, wemay regard G as a subset of G.


Claim: The induced topology on G as a subset of G is the Krull topologyon G.

Proof of claim: Let σUB be a basic open set in G. Let {β1, . . . , βk} be abasis for B over F. Then for τ ∈ G, τ ∈ σUB if and only if τ(βi ) = σ(βi ),i = 1, . . . , k. Thus σUB = G ∩ V where V = ∏

α∈E Sα with Sα = Rα forα �= βi and Sβi = {σ(βi )}, and so V is an open set in G.

Conversely, let V be a basic open set in G. Then V = ∏α∈E Sα with Sα =

Rα for all but finitely many α. Since each Rα is finite, we see that V is afinite union of open sets of the form V = ∏

α∈E Sα with Sα = Rα for all butfinitely many α, say for α �= α1, . . . , αk , and Sαi = {α′

i } is a single point foreach i = 1, . . . , k. Let B = F(α1, . . . , αk), a finite extension of F. If σ ∈ Gsatisfies σ(αi ) = α′

i , i = 1, . . . , k, then G ∩ V = σUB is open, while if thereis no σ ∈ G satisfying σ(αi ) = α′

i , i = 1, . . . , k, then G ∩ V is the empty set,which is also open.

Claim: G is a Hausdorff space.

Proof of claim: Let f1, f2 ∈ G with f1 �= f2. Then f1(α0) �= f2(α0) forsome α0 ∈ E (regarding G as a space of functions). Let A1 = ∏

α∈E Sα withSα = Rα for α �= α0 and Sα0 = { f1(α0)}, and let A2 = ∏

α∈E Sα with Sα = Rα

for α �= α0 and Sα0 = { f2(α0)}. Then f1 ∈ A1, f2 ∈ A2, and A1 and A2 aredisjoint open sets.

(Since G is a subspace of a Hausdorff space, it is Hausdorff, but for sim-plicity we chose to prove that directly in step (1). Observe that the proofs thatG and G are Hausdorff are very similar. In the same way, we may adapt theproof that G is totally disconnected to prove that G is totally disconnected.However, it is G and not G that we are interested in.)

Claim: G is a closed subset of G.

Proof of claim: Let f ∈ G, f /∈ G. We need to show f has an openneighborhood V that is disjoint from G. Note every f ∈ G satisfies f | F = id,so if f /∈ G, f is not an automorphism of E. If f : E → E is an injective mapof fields, then, by Lemma 5.4.13, f is an automorphism of E, contradictingour assumption. Thus there are two possibilities:

(1) f is not an injection; or(2) f is an injection but not a map of fields.In case (1) let α1, α2 ∈ E with f (α1) = f (α2), α1 �= α2. Let V = ∏

α∈E Sα

with Sα1 = { f (α1)}, Sα2 = { f (α2)}, and Sα = Rα for α �= α1, α2. Thenf ∈ V , V is open, and G ∩ V is the empty set (as any g ∈ G is an injection,so g(α1) �= g(α2)).

In case (2) we must have one of the following: f (α1 + α2) �= f (α1) +f (α2) for some α1, α2 ∈ E, f (α1α2) �= f (α1) f (α2) for some α1, α2 ∈ E,


or f (α−11 ) �= f (α1)

−1 for some 0 �= α1 ∈ E. For simplicity assume it isthe first of these. Let V = ∏

α∈E Sα with Sα1 = { f (α1)}, Sα2 = { f (α2)},Sα1+α2 = { f (α1 + α2)}, and Sα = Rα for α �= α1, α2, α1 + α2. Then f ∈ V ,V is open, and G ∩ V is the empty set (as any g ∈ G is a map of fields, sog(α1 + α2) = g(α1) + g(α2).

Claim: G is compact in the Krull topology.

Proof of claim: We merely need to assemble our previous work. The Krulltopology on G is the topology G inherits as a subspace of G. We have shownthat G is a compact Hausdorff space and that G is a closed subspace of G. Butit is a basic topological fact that a closed subspace of a compact Hausdorffspace is compact. ��Remark 5.4.16. While we do not need this for our purposes, it is worthwhileto put things in a larger context. An important topology on function spacesis the compact-open topology. If F = { f : X → Y } is the set of continuousfunctions from X to Y , the compact-open topology on F has as subbases thesets Z A,B = { f ∈ F | f (A) ⊆ B} where A ⊆ X is compact and B ⊆ Y isopen. In our case, E is discrete so every f : E → E is continuous, A ⊆ E iscompact if and only if it is finite, and every B ⊆ E is open. Now G ⊆ F andit is easy to see that the topology we have defined on G is the restriction of thecompact-open topology on F to G, i.e., that this topology is the compact-opentopology on G, and then we further see that the Krull topology on G agreeswith the compact-open topology on G. �

Let {Bi } be a set of fields intermediate between E and F. GeneralizingDefinition 3.4.1, we say that {Bi } are disjoint extensions of F if Bi ∩ Di = Ffor each i , where Di is the composite of the fields {B j } for j �= i . Then wehave the following generalization of Theorem 3.4.7.

Theorem 5.4.17. Let E be an extension of F and let {Bi ⊆ E} be disjointGalois extensions of F with Gi = Gal(Bi/F). Let B be the composite of {Bi }and let G = Gal(B/F). Then B is a Galois extension of F and G = ∏

i Gi

with the product topology.

Proof. A Zorn’s Lemma argument allows us to generalize Theorem 3.4.7 andconclude that B is a Galois extension of F and G = ∏

i Gi . We show G hasthe product topology.

Let U be a basic open neighborhood of the identity in the product topology.Then, by the definition of the product topology, U = ∏

i∈I Ui where Ui ⊆ Gi

is a basic open set and Ui = Gi except for finitely many i . Then, for eachi ∈ I , Ui = Gal(B/Ai ) where Ai is a finite extension of F, and Ai = F exceptfor finitely many i . Let A be the composite of the {Ai }. Then A is a finite

5.5 Exercises 171

extension of F, and U = UA is a basic open neighborhood of the identity inthe Krull topology on G.

Conversely, let U be a basic open neighborhood of the identity in the Krulltopology on G. Then, by definition, U = UA for some finite extension A of F.Let Ai = A ∩ Bi and note that, since {Bi } are disjoint extensions of F, Ai = Ffor all but finitely many i . Then, setting Ui = UAi , we have that each Ui isopen, Ui = Gi for all but finitely many i , and U = ∏

i∈I Ui is a basic openneighborhood of the identity in the product topology. ��Example 5.4.18. Here are two cases of infinite Galois extensions E of F andtwo subgroups H1 �= H2 of G = Gal(E/F) with Fix(H1) = Fix(H2). (Ofcourse, H1 = H2.)

(1) Let J = {−1, 2, 3, 5, . . . } consist of −1 and the primes. For eachj ∈ J , let B j = Q(

√j) ⊂ C, and let B be the composite of the {B j }. Note {B j }

are disjoint extensions of Q. Let G j = Gal(B j/Q). Note G j is isomorphic toZ/2Z for each j , so, by Theorem 5.4.17, G = Gal(B/Q) = ∏

j∈J (Z/2Z), andG is uncountable. Hence G has uncountably many subgroups of index 2. Butfor any subgroup H of index 2, (Fix(H)/Q) = 2 by Proposition 5.4.11. How-ever, Q has only countably many extensions of degree 2 (as each is obtainedby adjoining a root of a quadratic equation to Q, and there are only countablymany quadratic equations with rational coefficients). Hence there are two (andindeed, uncountably many) distinct subgroups H1 and H2 of G of index 2 withFix(H1) = Fix(H2).

(2) Fix a prime p and an algebraic closure Fp of Fp. Then Fp is the com-posite of the fields Bn = Fpn contained in Fp. Let G = Gal(Fp/Fp) and let ∈ G be the Frobenius map, (α) = α p. Let H be the subgroup of G gen-erated by . Then Fix(H) = Fp = Fix(G). We claim H �= G. To see this, letA be the composite of the fields Bn where n is a power of 2, and let D be thecomposite of the fields Bn for n odd. Then, by Lemma 3.4.2, A and D are dis-joint extensions of Fp, and AD = Fp. Hence G = Gal(A/Fp) × Gal(D/Fp).Let | A = A and | D = D. Then neither A nor D is the identitymap. Now H is a cyclic group, so H = {k} = {k

AkD}. But G also has the

subgroups {kA} and {k

D} (and uncountably many others), so H �= G. �

5.5 Exercises

Exercise 5.5.1. Let B1 and B2 be finite extensions of F satisfying property“P”. Show that B1B2 and B1 ∩ B2 also satisfy property “P”. Here “P” is anyone of the following: finite separable, finite purely inseparable, finite normal,separable, purely inseparable, normal, Galois. (“P” finite Galois is handled inCorollary 3.4.8 and in Exercise 3.9.13.)


Exercise 5.5.2. (a) Let A0 be the field obtained by adjoining all the complexroots of every irreducible polynomial f (X) ∈ Q[X ] of odd degree. For i > 0,let Ai be the field obtained from Ai−1 by adjoining all complex square rootsof all elements of Ai−1. Show that A0 ⊂ A1 ⊂ A2 ⊂ · · · and that

Q =∞⋃

i=0

Ai .

(b) Show that the conclusion of part (a) remains true if A0 is the field obtainedby adjoining a single real root of every irreducible polynomial f (X) ∈ Q[X ]of odd degree.

Exercise 5.5.3. Let Qsolv = {z ∈ C | mz(X) is solvable by radicals}. Let B0 =Q. For i > 0, let Bi be the field obtained from Bi−1 by adjoining all complexnth-roots of all elements of Bi−1, for all positive integers n. Show that B0 ⊂B1 ⊂ B2 ⊂ · · · and that

Qsolv =∞⋃

i=0

Bi .

Exercise 5.5.4. Recall that Q denotes the field of algebraic numbers. Showthat the only nontrivial element of finite order in Gal(Q/Q) is the automor-phism of Q given by complex conjugation (an element of order 2).

Exercise 5.5.5. Let F = C and let E = C(X). Note that E is not an algebraicextension of F. Let G = {σ : E → E | σ | F = id}. Show that Fix(G) =F. (This shows that in Section 5.4, it was necessary to require that E be analgebraic extension of F. Otherwise we would have nonalgebraic “Galois”extensions and the theory developed there would not hold.)

Exercise 5.5.6. In the notation of Example 5.4.18 (1), show that the closedsubgroups of G of index 2 are in 1–1 correspondence with the nonempty finitesubsets of J .

Exercise 5.5.7. Explicitly exhibit uncountably many subgroups of the groupG of Example 5.4.18 (2).

6

Transcendental Extensions

The preceding chapters have dealt (almost) exclusively with algebraic exten-sions of fields. In this chapter we present an introduction to the theory of tran-scendental extensions.

6.1 General Results

Definition 6.1.1. (1) Let E be an extension of F. Then α ∈ E is transcendentalover F if α is not a root of any polynomial f (X) ∈ F[X ].

(2) E is a transcendental extension of F (or E/F is transcendental) if someα ∈ E is transcendental over F. �

It is convenient to adopt the following nonstandard terminology.

Definition 6.1.2. Let E be an extension of F. Then E is a completely transcen-dental extension of F (or E/F is completely transcendental) if every α ∈ E,α �∈ F, is transcendental over F. �Lemma 6.1.3. Let E be any extension of F. Then there is a unique field B withF ⊆ B ⊆ E and with B algebraic over F and E completely transcendentalover B.

Proof. B = {α ∈ E | α is algebraic over F}. ��Definition 6.1.4. Let E be an extension of F. Then E is a purely transcendentalextension of F (or E/F is purely transcendental) if E is isomorphic to the fieldof rational functions F(X1, X2, . . . ) in the (finite or infinite) set of variables{X1, X2, . . . }. �


174 6 Transcendental Extensions

Lemma 6.1.5. Let E be a purely transcendental extension of F. Then E is acompletely transcendental extension of F.

Proof. We may assume that E = F(X1, X2, . . . ). Let α ∈ E. Then α =p(X1, . . . , Xk)/q(X1, . . . , Xk) for some polynomials p(X1, . . . , Xk) andq(X1, . . . , Xk). (Note that any rational function can only involve finitely manyvariables.) Suppose that α is algebraic over F. Then α is a root of some irre-ducible polynomial

∑ni=0 ai Xi in F[X ]. Substituting, and clearing denomina-

tors, we have a0q(X1, . . . , Xk)n+∑n−1

i=1 ai p(X1, . . . , Xk)i q(X1, . . . , Xk)

n−i+an p(X1, . . . , Xk)

n = 0. Now F[X1, X2, . . . , Xk] is a unique factorization do-main, and q(X1, . . . , Xk) divides every term in this expression except possiblythe last one, so q(X1, . . . , Xk) must divide p(X1, . . . , Xk)

n and hence alsop(X1, . . . , Xk); similarly p(X1, . . . , Xk) must divide q(X1, . . . , Xk). Henceα = p(X1, . . . , Xk)/q(X1, . . . , Xk) ∈ F. ��Remark 6.1.6. Of course, if F is algebraically closed then every extension ofF is completely transcendental. We will give examples of extensions of C thatare not purely transcendental in Section 6.3. �Definition 6.1.7. Let E be an extension of F. A subset S = {s1, s2, . . . } ofE is algebraically independent over F if f (s1, s2, . . . ) �= 0 for any nonzeropolynomial f (X1, X2, . . . ) in F[X1, X2, . . . ]. �Lemma 6.1.8. Let E be an extension of F and let S = {s1, s2, . . . } ⊂ E bealgebraically independent over F. Let B = F(S) = F(s1, s2, . . . ). Then B isisomorphic to F(X1, X2, . . . ). In particular, B is purely transcendental.

Proof. Let ϕ0 : F[X1, X2, . . . ] → F[s1, s2, . . . ] be defined by ϕ0(Xi ) = si

for each i . ϕ0 is obviously surjective. It is injective as S is algebraically inde-pendent over F, so it is an isomorphism. But then, again as S is algebraicallyindependent over F, ϕ0 extends to an isomorphism ϕ : F(X1, X2, . . . ) →F(s1, s2, . . . ). ��

We adopt the convention that if S is the empty set, then S is algebraicallyindependent and F(S) = F.

We now develop the theory of transcendence bases. The reader should notethe very strong analogy between the notion, and properties of, a transcendencebasis of a field extension and a basis of a vector space. We begin with a tech-nical lemma, which is of interest in itself.

Lemma 6.1.9. Let E be an extension of F and let S be a subset of E. Let s1 bean arbitrary element of S. Then S is algebraically independent if and only ifS′ = S −{s1} is algebraically independent and s is transcendental over F(S′).

6.1 General Results 175

Proof. We prove the contrapositive of the statement of the lemma.Clearly, if S′ is not algebraically independent or if s1 is algebraic over

F(S′), then S is not algebraically independent, so it remains to prove the re-verse implication.

Let S′ = {s2, s3, . . . } and let B = F(S′). Suppose that S is not algebraicallyindependent. Then there is some polynomial f (X1, X2, . . . ) with coefficientsin F with f (s1, s2, . . . ) = 0. If the variable X1 does not appear in this poly-nomial then f (s1, s2, . . . ) = g(s2, . . . ) and S′ is not algebraically indepen-dent. If the variable X1 appears in this polynomial then we gather terms inlike powers of X1 together, and substitute si for Xi for i > 1. Then, writingh(X1) = f (X1, s2, s3, . . . ) = ∑

hk(s2, s3, . . . )Xk1 ∈ B[X1], we have that

h(s1) = 0, so s1 is algebraic over S′. ��Definition 6.1.10. Let E be an extension of F and let S = {s1, s2, . . . } bea subset of E. S is a transcendence basis of E over F if S is algebraicallyindependent over F and E is an algebraic extension of F(S). �Remark 6.1.11. At first glance, it may seem more logical to require that E =F(S) in the definition of a transcendence basis. But if we were to make thatdefinition, by Lemma 6.1.8 only purely transcendental extensions would havetranscendence bases, and that would greatly restrict the utility of the theory.So we make the more general definition. �Theorem 6.1.12. Let E be an extension of F and let R ⊆ T be subsets ofE such that R is algebraically independent over F and E is an algebraicextension of F(T ). Then there is a transcendence basis S of E over F withR ⊆ S ⊆ T .

Proof. The proof is a Zorn’s Lemma argument. Let S be the set of subsets ofE defined by

S = {S | R ⊆ S ⊆ T and S is algebraically independent over F}.S is nonempty as R ∈ S. Partially order S by inclusion. Clearly every

totally ordered subset of S has a maximal element, its union. Thus, by Zorn’sLemma, S has a maximal element S. We claim that S is a transcendence basisfor E over F.

By definition, S is algebraically independent over F, so we need only showthat E is algebraic over B = F(S). Suppose not and let α ∈ E be transcendentalover B. Then, by Lemma 6.1.9, S = S ∪ {α} is algebraically independent,contradicting the maximality of S. ��


Corollary 6.1.13. Let E be an extension of F.(1) E has a transcendence basis over F.(2) If R is any subset of E that is algebraically independent over F, then E

has a transcendence basis over F that contains R.(3) If T is any subset of E with E an algebraic extension of F(T ), then E

has a transcendence basis over F that is contained in T .

Proof. For (1), let R = ∅ and T = E in Theorem 6.1.12. For (2), let T = E inTheorem 6.1.12. For (3), let R = ∅ in Theorem 6.1.12. ��

The following “replacement lemma” is the key step in the proof of Theo-rem 6.1.15.

Lemma 6.1.14. Let E be an extension of F and let S and T be any two tran-scendence bases of E over F. Let s be any element of S and let S′ = S − {s}.Then there is an element t of T such that S′′ = S′ ∪ {t} is a transcendencebasis of E over F.

Proof. By Lemma 6.1.9 we see that s is transcendental over F(S′). In partic-ular, E is transcendental over F(S′). This implies that some element t of T istranscendental over F(S′), as if not, F(T ) would be algebraic over F(S′). Byassumption, T is a transcendence basis for E over F, so in particular E is al-gebraic over F(T ). But E algebraic over F(T ) and F(T ) algebraic over F(S′)implies E algebraic over F(S′), a contradiction.

Since t is transcendental over S′, S′′ is algebraically independent, againby Lemma 6.1.9. Furthermore, s is algebraic over F(S′′) as otherwise S ∪ {t}would be algebraically independent, which it is not as E is algebraic over F(S).Thus F(S) is algebraic over F(S′′). Again, E is algebraic over F(S), and so Eis algebraic over F(S′′). Hence S′′ satisfies both conditions for a transcendencebasis of E over F. ��

In the following theorem, we do not distinguish between the cardinality ofinfinite sets.

Theorem 6.1.15. Let E be an extension of F. Then any two transcendencebases of E over F have the same number of elements.

Proof. To prove the theorem it suffices to show that if E has a transcendencebasis S over F with a finite number k of elements, then any subset T of Ewith more than k elements (perhaps with infinitely many elements) cannot bealgebraically independent. We prove this by contradiction.

Let S = {s1, s2, . . . , sk} and suppose that T is a transcendence basis havingmore than k elements. Apply Lemma 6.1.14 k times in succession. First we


obtain a transcendence basis S1 = {t1, s2, s3, . . . , sk} where t1 is some elementof T . Next we obtain a transcendence basis S2 = {t1, t2, s3, . . . , sk} wheret1 and t2 are some elements of T . Finally we obtain a transcendence basisSk = {t1, t2, t3, . . . , tk} where each ti is an element of T . But T has more thank elements, so there is an element tk+1 of T that is not in Sk . But then, byLemma 6.1.9, tk+1 is transcendental over F(Sk), contradicting the fact that Sk

is a transcendence basis for E over F. ��With this theorem in hand, we may make the following important defini-

tion.

Definition 6.1.16. Let E be an extension of F. Then tr-deg(E/F), the transcen-dence degree of E over F, is equal to the number of elements in any transcen-dence basis S of E over F, tr-deg(E/F) ∈ {0, 1, 2, . . . } ∪ {∞}. �Remark 6.1.17. If E = F(X1, X2, . . . ) then {X1, X2, . . . } is a transcendencebasis for E over F and so tr-deg(E/F) is the number of elements of this set.Thus we see that F(X1, X2, . . . , Xk) is not isomorphic to F(X1, X2, . . . , X�)

if k �= �. �Theorem 6.1.18. Let B be an extension of F with transcendence basis S andlet E be an extension of B with transcendence basis T . Then E is an extensionof F with transcendence basis S ∪ T . In particular,

tr-deg(E/F) = tr-deg(E/B) + tr-deg(B/F).

Proof. Let B0 = F(S), so B0 ⊆ B. Let E0 = B0(T ) and let E1 = B(T ), soE0 ⊆ E1 ⊆ E.

Since T is algebraically independent over B, it is certainly algebraicallyindependent over B0. This implies that S ∪T is algebraically independent overB. For suppose there were a nonzero polynomial f (X1, . . . , Y1, . . . ) with co-efficients in B with f (s1, . . . , t1, . . . ) = 0. By the algebraic independence ofT none of the variables Yi can appear, so f (s1, . . . , Y1, . . . ) = g(X1, . . . )

for some nonzero polynomial g(X1, . . . ) with coefficients in B. But theng(s1, . . . ) = 0, contradicting the algebraic independence of S.

Now F(S ∪ T ) = (F(S))(T ) = B0(T ) = E0. Let α ∈ E1 = B(T ). Bydefinition, α = p(T1, . . . )/q(T1, . . . ) for some polynomials with coefficientsin B. Let B1 be the extension of B0 obtained by adjoining the finitely manycoefficients of these polynomials, all of which are algebraic over B0 = F(T )

(as B is algebraic over B0, since S is a transcendence basis). Then α ∈ B1(T ),which is a finite extension of E0 = B0(T ), and so α is algebraic over E0. Sinceα ∈ E1 was arbitrary, this shows that E1 is algebraic over E0.


By the definition of a transcendence basis, E is algebraic over E1, and wehave just shown that E1 is algebraic over E0, so E is algebraic over E0 =F(S ∪ T ). Hence S ∪ T satisfies both conditions for a transcendence basis ofE over F.

Note that S and T are disjoint as S ⊂ B − F and T ⊂ E − B, so the secondpart of the theorem immediately follows. ��Corollary 6.1.19. Let E be any extension of F. Then there is a field B withF ⊆ B ⊆ E and with B purely transcendental over F and E algebraic over B.

Proof. Let S be a transcendence basis for E over F and let B = F(S). ��Remark 6.1.20. Comparing Corollary 6.1.19 with Lemma 6.1.3, we see that inthe situation of Corollary 6.1.19, the field B is not unique. �

Let B and D be two extensions of F, with B algebraic and D purely tran-scendental. Then Lemma 6.1.5 shows that B and D are disjoint extensions ofF. In fact, more is true.

Theorem 6.1.21. Let B and D be subfields of E containing F. Suppose that Bis an algebraic extension of F and that D is a purely transcendental extensionof F. Then B and D are linearly disjoint extensions of F.

Proof. We first prove this in the case that k = tr-deg(D/F) is finite. In thiscase let D = F(X1, . . . , Xk). We prove the theorem by induction on k.

To begin the induction, we let k = 1. Let {Y1, . . . , Y�} be a F-linearly in-dependent set of elements in F(X1), and suppose

∑�j=1 β j Y j = 0 with each

β j ∈ F, and not all of them 0. Clearing denominators, we may assume thateach Y j is a polynomial in X1. Performing a nonsingular change of basis (re-ordering and adding multiples of each of these polynomials to the others), ifnecessary, we may obtain a (necessarily F-linearly independent) set of nonzeropolynomials {Y ′

1, . . . , Y ′�} with deg(Y ′

1) > deg(Y ′2) > · · · > deg(Y ′

�), and0 = ∑�

j=1 β j Y j = ∑�j=1 β ′

j Y′j for uniquely determined elements β ′

j ∈ F, notall of which are 0. But this is a nontrivial polynomial relationship f (X1) = 0with f a polynomial with coefficients in B, and thus we see that X1 is alge-braic over B. But B is algebraic over F, so that implies that X1 is algebraicover F, a contradiction.

Now suppose the theorem is true for k − 1 and consider F(X1, . . . , Xk).Let {Y1, . . . , Y�} be a F-linearly independent set of elements in F(X1, . . . , Xk),and suppose

∑�j=1 β j Y j = 0 with β j ∈ F for each j , not all of which

are 0. By the truth of the k − 1 case, we cannot have {Y1, . . . , Y�} ⊆F(X1, . . . , Xk−1). Then, proceeding as in the k = 1 case, we obtain f (Xk) = 0


for some nontrivial polynomial f with coefficients in B(X1, . . . , Xk−1). ThenXk is algebraic over B(X1, . . . , Xk−1), and B(X1, . . . , Xk−1) is algebraic overF(X1, . . . , Xk−1), so Xk is algebraic over F(X1, . . . , Xk−1), a contradiction.

Finally, a linear dependence of a set of elements of F(X1, X2, . . . ) mustbe a linear dependence among elements of F(X1, . . . , Xk) for some k, so thefinite case implies the general case. ��Theorem 6.1.22. Let B and D be subfields of E containing F. Suppose that Band D are linearly disjoint extensions of F, and that D is a purely transcen-dental extension of F. Then BD is a purely transcendental extension of B andtr-deg(BD/B) = tr-deg(D/F).

Proof. Let D = F(X1, X2, . . . ). Clearly BD = B(X1, X2, . . . ) so to prove thetheorem we need only show that {X1, X2, . . . } is algebraically independentover B. Suppose the elements of this set satisfy some polynomial relation-ship. This can only involve finitely many variables, so we may assume it isf (X1, . . . , Xk) = 0 for some k, where f is a polynomial with coefficientsin B. Write this polynomial as a sum of terms, each of which is a distinctmonomial in {X1, . . . , Xk} times a coefficient in B. Now the distinct monomi-als in {X1, . . . , Xk} are linearly independent over F, as {X1, . . . , Xk} is alge-braically independent over F. By the definition of linear disjointness, each ofthe coefficients in f must be 0, and so f (X1, . . . , Xk) is the 0 polynomial, asrequired. ��

Here are a pair of examples that illustrate many of the concepts and resultsthat we have developed in this section.

Example 6.1.23. Let E = C(X). Note that {X}, {X2}, and {X3} are all tran-scendence bases for E over C. Set U = X2 and V = X3. Let B1 = C(U )

and B2 = C(V ). Note that E is a Galois extension of B1 of degree 2 withGalois group Gal(E/B1) generated by the automorphism of E determined byσ1(X) = −X . Also note that E is a Galois extension of B2 of degree 3 withGalois group Gal(E/B2) generated by the automorphism of E determined byσ2(X) = ωX . Clearly B1B2 = E as X = V/U ∈ B1B2. Note that σ1 andσ2 generate a group G of automorphisms of E of order 6. Indeed this is acyclic group with generator σ0 = σ1σ

−12 where σ0(X) = exp(2π i/6)X . Then

B0 = Fix(G) = B1 ∩ B2 = C(W ) where W = X6. In this case we have

tr-deg(B1/B0) = tr-deg(B2/B0) = 0, tr-deg(E/B1) = tr-deg(E/B2) = 0.

Observe that E = B1(V ) and V has minimal polynomial mV (Z) = Z2 −U 3 ∈ B1[Z ], and similarly that E = B2(U ) and U has minimal polynomialmU (Z) = Z3 − V 2 ∈ B2[Z ].


Finally, observe that in this case B1 and B2 are not disjoint extensionsof C. �Example 6.1.24. Let E = C(X). Note that {X}, {X2}, and {(X + 1)2} are alltranscendence bases for E over C. Set U = X2 and V = (X + 1)2. Let B1 =C(U ) and B2 = C(V ). Note that E is a Galois extension of B1 of degree 2 withGalois group Gal(E/B1) generated by the automorphism of E determined byσ1(X) = −X . Also note that E is a Galois extension of B2 of degree 2 withGalois group Gal(E/B2) generated by the automorphism of E determined byσ2(X) = −X − 2. Clearly B1B2 = E as X = (V − U − 1)/2 ∈ B1B2. Thegroup G of automorphisms of E generated by σ1 and σ2 contains the elementσ0 = σ2σ1 and σ0(X) = X + 2. Hence B0 = Fix(G) = B1 ∩ B2 = C as forno nonconstant rational function f (X) do we have f (X) = f (X + 2). In thiscase we have

tr-deg(B1/B0) = tr-deg(B2/B0) = 1, tr-deg(E/B1) = tr-deg(E/B2) = 0.

Observe that E = B1(V ) and V has minimal polynomial mV (Z) = Z2 −2(U + 1)Z + (U − 1)2 ∈ B1[Z ], and similarly that E = B2(U ) and U hasminimal polynomial mU (Z) = Z2 − 2(V + 1)Z + (V − 1)2 ∈ B2[Z ].

Finally, observe that in this case B1 and B2 are disjoint but not linearlydisjoint extensions of C. �

Let us now give an application of the theory we have just developed tothe theory of symmetric functions. Compare the proof of Lemma 3.1.12 andRemark 3.1.13. We adopt the notation of Section 3.1 here.

Theorem 6.1.25. Let D be an arbitrary field, let E = D(X1, . . . , Xd) be thefield of rational functions in the variables X1, . . . , Xd and let F ⊆ E be thefield of symmetric functions in X1, . . . , Xd. Let s1, . . . , sd be the elementarysymmetric polynomials. Then S = {s1, . . . , sd} is algebraically independent.Furthermore, F is a purely transcendental extension of D and S is a transcen-dence basis for F over D.

Proof. By definition, F is the fixed field of the symmetric group Sd acting onE by permuting the variables, so E is an algebraic extension of F; indeed, Eis a Galois extension of F with Galois group Gal(E/F) = Sd . We showed inLemma 3.1.4 that F = D(s1, . . . , sd). Thus it only remains to show that S ={s1, . . . , sd} is algebraically independent. Suppose not. By Corollary 6.1.13(3), S contains a proper subset R of cardinality c < d that is a transcendencebasis for F over D. But then

c = 0 + c = tr-deg(E/F) + tr-deg(F/D) = tr-deg(E/D) = d,

a contradiction. ��

6.2 Simple Transcendental Extensions 181

6.2 Simple Transcendental Extensions

Definition 6.2.1. A simple transcendental extension of the field F is a field Eobtained by adjoining a single transcendental element X to F, E = F(X). �

Throughout this section we let E = F(X).A simple transcendental extension E is easy to describe: it is a purely tran-

scendental extension of transcendence degree 1. But there are interesting ques-tions we can ask about E.

Lemma 6.2.2. Let Y = p(X)/q(X) ∈ E, with p(X) and q(X) relativelyprime polynomials in F[X ] of degrees d1 and d2 respectively, and let B =F(Y ). Suppose that d > 0. Let d = max(d1, d2). Then (E/B) = d.

Proof. (1) Since E = F(X), certainly E = B(X). We prove the lemma byshowing that X satisfies an irreducible polynomial of degree d over B.

Evidently X is a root of the polynomial f (Y, Z) = Y q(Z)− p(Z) ∈ B[Z ].Thus we need only show that this polynomial is irreducible.

It is easy to see that f (Y, Z) is irreducible in (F[Z ])[Y ]: f (Y, Z) islinear in Y so any factorization would have to be of the form f (Y, Z) =a(Z)(b(Z)Y + c(Z)). Then q(Z) = a(Z)b(Z) and −p(Z) = a(Z)c(Z),and hence a(Z) is a common factor of p(Z) and q(Z). But p(Z) and q(Z) areassumed relatively prime so a(Z) is a unit, i.e., a constant polynomial.

Now (F[Z ])[Y ] = F[Z , Y ] = F[Y, Z ] = (F[Z ])[Y ] so f (Y, Z) is irre-ducible in (F[Z ])[Y ]. We need to show it is irreducible in B[Y ] = (F(Z))[Y ].But that is an immediate consequence of Gauss’s Lemma, Lemma 4.1.4. (Weproved Gauss’s Lemma for Z and Q, but the exact same proof works for anyunique factorization domain R and its quotient field.) ��

We now determine the automorphism group of a simple transcendentalextension. Recall that GL2(F) denotes the group of invertible 2 × 2 matriceswith coefficients in the field F. It is easy to see that its center Z = Z(GL2(F))

is the subgroup consisting of all nonzero scalar multiples of the identity matrix(and so is naturally isomorphic to the multiplicative group F∗). As usual, welet PGL2(F) be the quotient PGL2(F) = GL2(F)/Z . PGL2(F) is known asthe projective linear group of degree 2 over F.

Theorem 6.2.3. Let AutF(E) be the group of automorphisms of E that fix F.

(1) For a matrix M =[

a bc d

]∈ GL2(F), let ϕM : E → E be defined by

ϕM( f (X)) = f

(aX + b

cX + d

).


Then ϕM ∈ AutF(E). Furthermore, any ψ ∈ AutF(E) is ψ = ϕM for someM ∈ GL2(F).

(2) The map : GL2(F) → AutF(E) given by (M) = ϕM−1 induces anisomorphism

: PGL2(F) → AutF(E).

Proof. (1) Observe that the polynomials aX + b and cX + d are relativelyprime if and only if det(M) = ad − bc �= 0. Assuming that, let Y = ψ(X) =(a X + b)/(cX + d). Since E = F[X ], ψ extends to a unique homomorphismfrom E to itself, and that homomorphism is ϕM . By Lemma 6.2.2, F(Y ) =E, so ϕM is surjective, and it is clearly injective, so it is an automorphism.Furthermore, again by Lemma 6.2.2, if ψ is any automorphism of E, thenY = ψ(X) must be of the above form, i.e., Y = ϕM(X) for some invertiblematrix M , in which case ψ = ϕM .

(2) By (1), is a surjection of sets. We now check that is a homo-morphism, i.e., that (M1 M2) = (M1)(M2). This is a direct computation,using the fact that ψ ∈ AutF(E) is determined by ψ(X):

Let M−1i =

[ai bi

ci di

]for i = 1, 2. First note that

(M1 M2)−1 = M−1

2 M−11 =

[a2 b2

c2 d2

] [a1 b1

c1 d1

]=

[a2a1 + b2c1 a2b1 + b2d1

c2a1 + d2c1 c1b1 + d2d1

].

Then, noting that for any automorphism ψ and any rational function f (X),the value of ψ on f (X) ∈ E is f (ψ(X)), so that the order of composition isreversed,

((M1)(M2))(X) = (ϕM−11

ϕM−12

)(X) = ϕM−12

(ϕM−11

(X))

= ϕM−12

(a1 X + b1

c1 X + d1

)

= a2a1 X+b1c1 X+d1

+ b2

c2a1 X+b1c1 X+d1

+ d2

= a2(a1 X + b1) + b2(c1 X + d1)

c2(a1 X + b1) + d2(c1 X + d1)

= (a2a1 + b2c1)X + (a2b1 + b2d1)

(c2a1 + d2c1)X + (c2b1 + d2d1)

= ϕM−12 M−1

1(X) = ϕ(M1 M2)−1(X) = (M1 M2)(X)

so is an epimorphism of groups, and hence induces an isomorphism ofgroups : GL2(F)/ Ker() −→ AutF(E). Let M be a matrix with M−1 =

6.2 Simple Transcendental Extensions 183[

a bc d

]. (M) = ϕM−1 is the identity on E if and only if ϕM−1(X) = X , i.e., if

and only if (a X + b)/(cX + d) = X , which occurs if and only if a = d �= 0and b = c = 0, i.e., if and only if M−1 is a nonzero scalar matrix, and this istrue if and only if M is a nonzero scalar matrix. ��Remark 6.2.4. (1) If you are familiar with fractional linear transformations,you will have recognized them in the statement of Theorem 6.2.3. But you mayhave been puzzled by the appearance of the matrix inverse in that statement.The group PGL2(R) acts on the upper half-plane H = {z ∈ C | Im(z) > 0}by fractional linear transformations (or Mobius transformations), ρM(z) =(az +b)/(cz +d), where the matrix M is as in the statement of Theorem 6.2.3.In that action we have ρM1 M2 = ρM1ρM2 , where on the right-hand side of thisequation multiplication is simply composition of maps from H to itself. Butthat action is not the action of PGL2(F) on E; as we remarked in the proofof Theorem 6.2.3, in this action the order of composition is reversed. (Theautomorphism ψ does not act on the rational function f (X) by acting on thevalue of the function f on the variable X , but rather by acting on the variableX which is the argument to the function f , and this accounts for the reversalof order.)

(2) Evidently ψ defined by ψ(X) = aX + b, a �= 0, defines an automor-phism of E fixing F, as does ψ(X) = 1/X . Thus AutF(E) must contain thegroup generated by compositions of these automorphisms as a subgroup. It iseasy to check that this subgroup is precisely the group of all fractional lineartransformations. Thus, Theorem 6.2.3 says that all automorphisms of E fixingF are obtained in this way.

Theorem 6.2.5 (Luroth). Let F ⊂ B ⊆ E. Then B is a simple transcendentalextension of F, i.e., B = F(Y ) for some Y ∈ E.

Proof. Let W ∈ B, W �∈ F. Then W = a(X)/b(X) for some relativelyprime polynomials a(X) and b(X) not both of which are constant, so, byLemma 6.2.2, the element X of E is algebraic over F(W ). Since F(W ) ⊆ B,X is algebraic over B. (This fact is also a consequence of Lemma 6.1.5 andTheorem 6.1.17, but is just as easy to prove directly.) Let X have minimumpolynomial m X (Z) over B,

m X (Z) = Zn + Yn−1 Zn−1 + · · · + Y0 with each Yi ∈ B.

Then (E/B) = n. Since X is not algebraic over F some coefficient Y = Yi0 ofm X (Z) is not in F. We claim that B = F(Y ).

Let B0 = F(Y ). Since Y ∈ B, B0 ⊆ B. Write Y as a rational function Y =p(X)/q(X) in lowest terms. Then, by Lemma 6.1.2, we have that (E/B0) = d


where d is the maximum of the degrees of p(X) and q(X). Then

d = (E/B0) = (E/B)(B/B0) = n(B/B0)

so if we show that n = d, then B = B0 = F(Y ) as required.Now X is a root of f (Z) = Y q(Z) − p(Z), so m X (Z) divides f (Z) in

B[Z ]. Note that f (Z) is not the 0 polynomial as if it were, we would haveY = p(Z)/q(Z) in B(Z), which is impossible as Y ∈ B and p(Z)/q(Z) �∈ B.Let f (Z) = m X (Z)g(Z). We may “clear denominators” in m X (Z) to obtain apolynomial

m(X, Z) = cn(X)Zn + cn−1(X)Zn−1 + · · · + c0(X)

with {cn(X), . . . , c0(X)} a relatively prime set of polynomials in F[X ], so thatm(X, Z) is a primitive polynomial of degree n when regarded as a polynomialin Z . By the definition of Y , we see that p(X) divides ci0(X) and q(X) dividescn(X), so, if m(X, Z) is regarded as a polynomial in X , its degree is at least d.For clarity we will write degZ (m(X, Z)) = n and degX (m(X, Z)) ≥ d.

Now f (Z) = m X (Z)g(Z), i.e.,

Y q(Z) − p(Z) = m X (Z)g(Z).

Multiplying by cn(X), we obtain the equation

cn(X)(Y q(Z) − p(Z)) = m(X, Z)g(Z),

a polynomial identity in B[X, Z ]. Substituting Y = p(X)/q(X) and multiply-ing by q(X), we obtain the equation

cn(X)(p(X)q(Z) − q(X)p(Z)) = m(X, Z)g(Z)q(X),

a polynomial identity in F[X, Z ]. Now cn(X) and m(X, Z) are relativelyprime, so m(X, Z) must divide p(X)q(Z) − q(X)p(Z), i.e.,

r(X, Z) = (p(X)q(Z) − q(X)p(Z)) = m(X, Z)s(X, Z)

for some polynomial s(X, Z) ∈ F[X, Z ].Now degX (r(X, Z)) = degZ (r(X, Z)) ≤ d. But we observed above

that degX (m(X, Z)) ≥ d, so we must have degX (m(X, Z)) = d. ThendegX (s(X, Z)) = 0, so s(X, Z) = t (Z) for some polynomial t (Z) ∈ F[Z ].As we have observed, m(X, Z) is primitive when regarded as a polynomial inZ , i.e., the coefficients of the different powers of Z are relatively prime poly-nomials in X . But then the same is true for r(X, Z) = m(X, Z)t (Z). But by

6.3 Plane Curves 185

symmetry, r(X, Z) is then also primitive when regarded as a polynomial in X ,so the coefficients of the different powers of X are relatively prime polynomi-als in Z , and hence t (Z) is a constant polynomial. Thus r(X, Z) = am(X, Z)

for some a ∈ F. We then conclude that

n = degZ (m(X, Z)) = degZ (r(X, Z)) = degX (r(X, Z)) = d,

as required. ��

6.3 Plane Curves

Throughout this section, except when explicitly stated otherwise, we let F = C,the field of complex numbers.

Hitherto we have only considered purely transcendental extensions. In thissection we investigate extensions that are not (necessarily) purely transcen-dental. Our examples are chosen from elementary algebraic geometry. Clearly,they are just the tip of an iceberg, but it is beyond our scope here to investigatethis subject more broadly or deeply.

Definition 6.3.1. Let f (X, Y ) ∈ F[X, Y ] be an irreducible polynomial. LetV ( f ) = {(z1, z2) ∈ F2 | f (z1, z2) = 0}. Then C = V ( f ) is the plane curveassociated to f (X, Y ). The quotient field of F[X, Y ]/〈 f (X, Y )〉 is the functionfield of C. �Remark 6.3.2. (1) Strictly speaking, we should call C as in Definition 6.3.1 anirreducible affine plane curve, but we adopt the shorter language.

(2) In Definition 6.3.1 we are using some commutative ring theory. In thisdefinition, 〈 f (X, Y )〉 denotes the ideal in F[X, Y ] generated by the polynomialf (X, Y ). Since F[X, Y ] is a unique factorization domain, and we are assumingf (X, Y ) is irreducible, then f (X, Y ) is prime and 〈 f (X, Y )〉 is a prime ideal.Thus F[X, Y ]/〈 f (X, Y )〉 is an integral domain and we may indeed take itsquotient field.

(3) We are restricting our attention to C[X, Y ], rather than, for example,Q[X, Y ], for the following reasons: f (X, Y ) = X2 + Y 2 + 1 is an irreduciblepolynomial in Q[X, Y ], but V ( f ) = ∅ when we consider it over Q, and wedon’t want to regard that as a curve. Also, if f (X, Y ) and g(X, Y ) are re-garded as polynomials over C, then V ( f ) = V (g) if and only if f (X, Y )

and g(X, Y ) are constant multiples of each other, but this is not true overQ, as we see from the examples f (X, Y ) = X2 + Y 4 + 1 and g(X, Y ) =X4+Y 2+1. However, there are some (very!) interesting questions about V ( f )


when f (X, Y ) is regarded as a polynomial over Q, or over some field inter-mediate between Q and C, and we will return to this point in Remark 6.3.13below.

(4) Much of the discussion here is valid with C replaced by an algebraicallyclosed field of arbitrary characteristic. We are restricting our attention to char-acteristic 0 for simplicity. �Notation 6.3.3. In the situation of Definition 6.3.1, we write C(x, y) for thequotient field of the curve C, where, in the natural map from C[X, Y ] to thisquotient field, the image of X is x and the image of Y is y. �

Remark 6.3.4. Our notation C(x, y) is perfectly consistent with our previousnotation. C(x, y) is an extension field of C obtained by adjoining the elementsx and y. Also, this field C(x, y) is an extension of C of transcendence de-gree 1. If we assume that f (X, Y ) involves both X and Y (i.e, that it is not apolynomial in X alone, or a polynomial in Y alone), then each of x and y istranscendental over C; moreover, in this case, {x} and {y} are each transcen-dence bases for C(x, y) over C. Furthermore, in this case, the natural mapsfrom C(X) to C(x), and from C(Y ) to C(y), are field isomorphisms. �Definition 6.3.5. In the situation of Definition 6.3.1, the curve V ( f ) is a ra-tional curve if its function field C(x, y) is rational. �

We now present some interesting examples of plane curves that are or arenot rational.

Theorem 6.3.6. Fix a positive integer n and consider the Fermat curve Cn

given by Xn + Y n = 1.(1) For n = 2, Cn is rational.(2) For n ≥ 3, Cn is not rational.

Proof. (1) We begin by recalling the polynomial identity (T 2 −1)2 + (2T )2 =(T 2 + 1)2 which immediately yields the identity among rational functions

(T 2 − 1

T 2 + 1

)2

+(

2T

T 2 + 1

)2

= 1.

Guided by this identity, we construct an embedding ϕ : F(x, y) −→ F(t)defined by

ϕ(x) = (t2 − 1)/(t2 + 1) and ϕ(y) = 2t/(t2 + 1).


We thus see that F(x, y) is isomorphic to a subfield of F(t). In fact, thissubfield is F(t) itself. To show this, we need only show that ϕ is an epimor-phism. To see this, observe that ϕ(x)/ϕ(y) = (t2 − 1)/(2t). Cross-multiply,regard this as a quadratic equation in t , and solve by the quadratic formula, us-ing the relation ϕ(x)2 +ϕ(y)2 = 1, to obtain t = (ϕ(x)+1)/ϕ(y). (Of course,once we know that F(x, y) is isomorphic to a subfield of F(t), Luroth’s theo-rem (Theorem 6.2.5) tells us that F(x, y) is purely transcendental, but not onlyis the above argument simpler and more direct, it also gives us more informa-tion.)

(2) The isomorphism in part (1) was due to the polynomial identity amongsquares that we began with in that case. As we shall see, the fact that there isno isomorphism for n ≥ 3 is due to the fact that there is no similar polynomialidentity.

Suppose F(x, y) were purely transcendental. Then we would have an iso-morphism ϕ : F(x, y) −→ F(t). Then ϕ(x) = f1(t)/ f2(t) where f1(t)and f2(t) are relatively prime polynomials in F[t], and similarly ϕ(y) =g1(t)/g2(t) where g1(t) and g2(t) are relatively prime polynomials in F[t],and furthermore (

f1(t)

f2(t)

)n

+(

g1(t)

g2(t)

)n

= 1.

In the remainder of this proof we abbreviate f1(t) by f1, etc. Multiplyingthrough by ( f2g2)

n we obtain the polynomial identity

an + bn = cn

where a = f1g2, b = g1 f2, and c = g1g2 are polynomials in F[t], not all ofwhich are constant. We show this identity has no solutions.

We prove this by contradiction. Assume there is at least one solution, andchoose a solution where dmax = max(deg a, deg b, deg c) is as small as possi-ble. In this case a, b, and c are certainly pairwise relatively prime. Rewrite thisidentity as

an = cn − bn =n∏

k=1

(c − λkb),

where λ = exp(2π i/n). Note that at most one of the factors on the right-hand side of this equation can be a constant polynomial. We now use the factthat F[t] is a unique factorization domain. These factors are pairwise relativelyprime, as if c − λi b and c − λ j b had a common factor for some i �= j , then,taking linear combinations, a and b would have that common factor. Hence


each of these factors is itself an nth power, c − λi b = eni , i = 1, . . . , n. In

particular, since n ≥ 3, this is true for i = 1, 2, 3. Then simple algebra shows

en3 = c − λ3b

= −λ(c − λb) + (1 + λ)(c − λ2b)

= −λen1 + (1 + λ)en

2 .

Note that 1 + λ �= 0 as n �= 2. Then, choosing complex numbers α and β withαn = −λ and βn = 1 + λ, and setting a′ = αe1, b′ = βe2, and c′ = e3, wehave

(a′)n + (b′)n = (c′)n

with d ′max = max(deg a′, deg b′, deg c′) ≤ dmax/n, contradicting the minimal-

ity of dmax. ��Theorem 6.3.7. Fix positive integers n ≥ 3 and m ≥ 3. Let h(X) be anypolynomial of degree m with distinct roots and let C be the curve defined byY n = h(X). Then C is not rational.

Proof. The proof of this is an adaptation of the proof of Theorem 6.3.6(2).Let h(X) have roots λ1, . . . , λm . Then C is given by

Y p = cq(X − λ1) · · · (X − λm)

Replacing Y by Y cm1/n if necessary, we may assume h(X) is monic, and

we do so henceforth. Assume that C is rational. Then we have a pair f1(t) andf2(t) of relatively prime polynomials in F[t], and also a pair g1(t) and g2(t)of relatively prime polynomials in F[t], with

(g1

g2

)n

=(

f1

f2− λ1

)· · ·

(f1

f2− λm

).

Multiplying this equation by gn2 f m

1 we obtain the polynomial identity

gn1 f m

2 = gn2 ( f1 − λ1 f2) · · · ( f1 − λm f2).

Now gn2 must divide f m

2 , as g2 is relatively prime to g1, and f m2 must divide

gn2 as f2 is relatively prime to f1, which implies that f2 is relatively prime to

each term ( f1 − λi f2). Hence gn2 and f m

2 must be equal up to a unit factor,which we may again assume is 1, by an appropriate linear change of variableif necessary. Thus we obtain the identity

gn1 = ( f1 − λ1 f2) · · · ( f1 − λm f2).


The terms on the right-hand side are pairwise relatively prime so each mustbe an nth power. Let ( f1 − λi f2) = en

i for i = 1, 2, 3. The three polynomials( f1 − λ1 f2), ( f1 − λ2 f2), and ( f1 − λ3 f2) are linearly dependent, and notwo are multiples of each other, so there is a linear dependence between themwith each of the three coefficients nonzero. By an appropriate linear change ofvariable we may assume this linear dependence is en

1 + en2 − en

3 = 0, and wehave seen in the proof of Theorem 6.3.6(2) that this is impossible. ��

In order to handle the remaining cases we first prove a technical lemma.

Lemma 6.3.8. There does not exist a pair of polynomials ( f (X), g(X)) inF[X ], not both of which are constant, such that ai f (X) + bi g(X) is a squarein F[X ], i = 1, . . . , 4, where (ai , bi ) are pairs of elements in F, not both 0,satisfying the condition that (a j , b j ) �= (eai , ebi ) for any e in F and any j �= i .

Proof. Assume there exists such a pair ( f (X), g(X)), which we abbreviate as( f, g). Choose such a pair with max(deg( f ), deg(g)) minimal. Then certainlyf and g are relatively prime. This implies that ai f + bi g and a j f + b j g arerelatively prime for i �= j . Let p = a1 f + b1g and q = a2 f + b2g. Then thereare unique pairs (a′

i , b′i ), with a′

i p + b′i q = ai f + bi g, i = 3, 4. Now

4∏i=1

(ai f + bi g) = pq(a′3 p + b′

3q)(a′4 p + b′

4q).

The left-hand side of this equation is a product of squares, hence a square.Our conditions on (ai , bi ) ensure that the terms on the right-hand side of thisequation are pairwise relatively prime, and hence each of them must be asquare. Let p = r2 and q = s2. Then, with any arbitrary but fixed choiceof square roots,

a′i p + b′

i q = a′i r

2 + b′i s

2 =(√

a′i r +

√−b′

i s)(√

a′i r −

√−b′

i s), i = 3, 4.

Now each of these products is a square, and the factors are again relativelyprime, so they must each be squares. Thus, replacing f by r , g by s, and(a1, b1), . . . , (a4, b4) with (c1, d1), . . . , (c4, d4) where

(c1, d1) =(√

a′3,

√−b′

3

),

(c2, d2) =(√

a′3, −

√−b′

3

),

(c3, d3) =(√

a′4,

√−b′

4

),

(c4, d4) =(√

a′4, −

√−b′

4

),


we have a pair of polynomials (r, s) satisfying the given conditions withmax(deg(r), deg(s)) < max(deg( f ), deg(g)), a contradiction. ��Remark 6.3.9. Lemma 6.3.8 is sharp. If f (X) = X4 + 1 and g(X) = X2, thenai f (X) + bi g(X) is a square for (a1, b1) = (1, 2), (a2, b2) = (1, −2), and(a3, b3) = (0, 1).

Theorem 6.3.10. Fix a positive integer m ≥ 3. Let h(X) be any polynomialof degree m with distinct roots and let C be the curve defined by Y 2 = h(X).Then C is not rational.

Proof. We begin by following the proof of Theorem 6.3.7, whose notation weadopt. Exactly as in that proof, we obtain an identity

g21 = ( f1 − λ1 f2) · · · ( f1 − λm f2),

with the terms on the right-hand side pairwise relatively prime, and hence eacha perfect square. In case m ≥ 4, this is impossible by Lemma 6.3.3. In casem = 3, we have only 3 perfect squares. But in this case, the equation g2

2 = f 32

implies that f2 is a perfect square as well, so we may again apply Lemma 6.3.8.��

There is one case left open, which we now dispose of.

Theorem 6.3.11. Let h(X) be any polynomial of degree 2 with distinct rootsand let C be the curve defined by Y 3 = h(X). Then C is not rational.

Proof. Let h(X) = a X2 + bX + c. Then Y 3 = aX2 + bX + c = a(X +b/(2a))2 + (c − b2/(4a)) so (X + b/(2a))2 = Y 3/a − (c/a − b2/(4a2)), andby making a linear change of variable and interchanging X and Y we are backin the case of Theorem 6.3.10. ��Remark 6.3.12. Note that the proof of Theorem 6.3.6(1) holds over Q, fromwhich we may conclude that the curve X2 + Y 2 = 1 is rational over everyfield of characteristic 0. Also note that the proofs of Theorem 6.3.6(2), Theo-rem 6.3.7, Theorem 6.3.10, and Theorem 6.3.11 hold over every algebraicallyclosed field of characteristic 0, from which we may conclude that these curvesare not rational over any field of characteristic 0.

Remark 6.3.13. Now let F be a subfield of C. For f (X, Y ) ∈ F[X, Y ], it makessense to consider not only VC( f ) = {(z1, z2) ∈ C2 | f (z1, z2) = 0} but alsoVF( f ) = VC( f ) ∩ F2 = {(z1, z2) ∈ F2 | f (z1, z2) = 0}. Questions aboutVF( f ) are questions of great arithmetic (i.e., number-theoretic) interest. Forexample, let F = Q and consider f (X, Y ) = Xn +Y n −1 as in Theorem 6.3.6.

6.4 Exercises 191

By clearing denominators, any point (a/c, b/c) ∈ VQ( f ) gives us integers a,b, and c satisfying the equation an + bn = cn , and conversely.

If n = 2, then a, b, and c form a Pythagorean triple. Setting t = j/k in theproof of Theorem 6.3.6 and clearing denominators, this gives us a parameter-ization of Pythagorean triples by a = j2 − k2, b = 2 jk, c = j2 + k2. Withthe restrictions j > k > 0 and j and k relatively prime and not both odd, thisyields all primitive Pythagorean triples, without duplication. (A Pythagoreantriple is primitive if a, b, and c are relatively prime.) This can all be found inEuclid.

On the other hand, if n ≥ 3, Fermat claimed that this equation has nointeger solutions with a, b, and c all nonzero. This is the famous Fermat’slast theorem. This claim was made by Fermat in the middle of the seventeenthcentury and remained open for over 300 years, until being proved by Wilesat the end of the twentieth century. It is because of this that the curves Cn arecalled Fermat curves. �

6.4 Exercises

Exercise 6.4.1. Let x and y be algebraically independent elements of a tran-scendental extension E of F. Show that F(x) and F(y) are linearly disjointextensions of F.

Exercise 6.4.2. Consider the polynomial f (X, Y ) = X2 − 2XY + Y 2 − 2 ∈Q[X, Y ]. Show that f (X, Y ) is irreducible. Show that {(r1, r2) ∈ Q2 |f (r1, r2) = 0} = ∅. Let E = Q(x, y) be the function field of the “curve”f (X, Y ) = 0. Show that E = Q(

√2)(x). In particular, E is a transcendental

but not completely transcendental extension of Q.

Exercise 6.4.3. Let f (X, Y ) ∈ F[X, Y ] be an irreducible polynomial in whichboth X and Y appear. Show that x and y are both transcendence bases for thefunction field of the curve f (X, Y ) = 0. (Hence this field has transcendencedegree 1 over F.)

Exercise 6.4.4. (a) Let E be a purely transcendental extension of F and let Bbe an algebraic extension of F. Show that BE is an algebraic extension of Eand that (BE/E) = (B/F).

(b) Let E be a purely transcendental extension of F and let B be a Galoisextension of F. Show that BE is a Galois extension of E and that Gal(BE/E) =Gal(B/F).


Exercise 6.4.5. An extension E of F is a finitely generated extension of F if Eis obtained by adjoining finitely many elements (algebraic or transcendental)to F. Show that E is finitely generated over F if and only if k = tr-deg(E/F)

is finite and for some (and hence for any) transcendence basis X1, . . . , Xk ofE over F, E is a finite extension of F(X1, . . . , Xk).

Exercise 6.4.6. Let F ⊆ B ⊆ E. Show that E is finitely generated over F ifand only if E is finitely generated over B and B is finitely generated over F.

Exercise 6.4.7. Let D2n , the dihedral group of order 2n, be given by D2n =〈a, b | an = 1, b2 = 1, ba = a−1b〉. Then D2n acts on C(X) where theaction is determined by a(X) = exp(2π i/n)X and b(X) = 1/X . Show thatFix(D2n) = C(T ) where T = Xn + X−n .

Exercise 6.4.8. (a) Let σ1(X) = 1/X and let σ2(X) = 1−X . Show that σ1 andσ2 generate a subgroup G of Aut(C(X)/C) that is isomorphic to the dihedralgroup D6. (Considering the matrices that give σ1 and σ2, as in Theorem 6.2.3,will make this computation easier.)(b) Show that Fix(G) = C(T ) where T = (X2 − X + 1)3/(X2(X − 1)2).

Exercise 6.4.9. (a) Let B1 and B2 be extensions of F. Show thattr-deg(B1B2/B1) ≤ tr-deg(B2/F) and that tr-deg(B1B2/F) ≤ tr-deg(B1/F) +tr-deg(B2/F). (Note that Example 6.1.24 gives an example of disjoint exten-sions B1 and B2 where we have strict inequality.)(b) Let B1 and B2 be linearly disjoint extensions of F. Show thattr-deg(B1B2/B1) = tr-deg(B2/F) and that tr-deg(B1B2/F) = tr-deg(B1/F) +tr-deg(B2/F).

Exercise 6.4.10. (a) Let C be the curve defined by Y 2 = X2 + X3. Show thatC is rational over Q (and hence over any field of characteristic 0).(b) Let C be the curve defined by Y 2 = X2 − X4. Show that C is rational overQ (and hence over any field of characteristic 0).

Exercise 6.4.11. Let f (X, Y ) = aX2 + bXY + cY 2 + d X + eY + f be anarbitrary irreducible polynomial, and let C be the curve defined by f (X, Y ) =0. Show that C is rational over C.

Exercise 6.4.12. Show that Theorem 6.3.7 and Theorem 6.3.10 hold for anypolynomial h(X) with at least 3 roots of multiplicity prime to n, in the case ofTheorem 6.3.7, or to 2, in the case of Theorem 6.3.10.

Exercise 6.4.13. Determine conditions on p so that each case of Theorem6.3.6, Theorem 6.3.7, Theorem 6.3.10, Theorem 6.3.11, Exercise 6.4.10, andExercise 6.4.11 holds when F is an algebraically closed field of characteris-tic p.

6.4 Exercises 193

Exercise 6.4.14. Let F be a field of characteristic p, let E ⊃ F be a purelytranscendental extension of F, and let E′ ⊇ E be a finite extension of E. Showthat E′ is not a perfect field.

A

Some Results from Group Theory

A.1 Solvable Groups

Definition A.1.1. A composition series for a group G is a sequence of sub-groups G = G0 ⊇ G1 ⊇ . . . Gk = {1} with Gi a normal subgroup of Gi−1 foreach i . A group G is solvable if it has a composition series with each quotientGi/Gi−1 abelian. �Example A.1.2. (1) Every abelian group G is solvable (as we may chooseG0 = G and G1 = {1}).

(2) If G is the group of Lemma 4.6.1, G = 〈σ, τ | σ p = 1, τ p−1 = 1,τστ−1 = σ r 〉 where r is a primitive root mod p, then G is solvable: G =G0 ⊃ G1 ⊃ G2 = {1} with G1 the cyclic subgroup generated by σ . Then G1

is a normal subgroup of G with G/G1 cyclic of order p − 1 and G1 cyclic oforder p.

(3) If G is the group of Lemma 4.6.3, G = 〈σ, τ | σ 4 = 1, τ 2 = 1,τστ−1 = σ 3〉 , then G is solvable: G = G0 ⊃ G1 ⊃ G2 = {1} with G1

the cyclic subgroup generated by σ . Then G1 is a normal subgroup of G withG/G1 cyclic of order 2 and G1 cyclic of order 4.

(4) If G is a p-group, then G is solvable. This is Lemma A.2.4 below.(5) If G = Sn , the symmetric group on {1, . . . , n}, then G is solvable

for n ≤ 4. S1 is trivial. S2 is abelian. S3 has the composition series S3 ⊃A3 ⊃ {1}. S4 has the composition series S4 ⊃ A4 ⊃ V4 ⊃ {1} where V4 ={1, (12)(34), (13)(24), (14)(23)}. (Here An denotes the alternating group.)

(6) If G = Sn or G = An , then G is not solvable for n ≥ 5. This isCorollary A.3.6 below. �Lemma A.1.3. A finite group G is solvable if and only if it has a compositionseries satisfying any one of the following conditions:

196 A Some Results from Group Theory

(1) Gi−1/Gi is solvable for each i .(2) Gi−1/Gi is abelian for each i .(3) Gi−1/Gi is cyclic for each i .(4) Gi−1/Gi is cyclic of prime order for each i .

Proof. If Gi−1/Gi is solvable then we may “refine” the original sequence toG ⊇ · · · ⊇ Gi−1 = H0 ⊇ H1 ⊇ · · · ⊇ Hm = Gi ⊇ · · · ⊇ Gk = {1} with Hj

normal in Hj−1 and Hj−1/Hj abelian; similarly we may refine a sequence withabelian quotients to one with cyclic quotients and one with cyclic quotients toone with quotients cyclic of prime order. On the other hand, if Gi−1/Gi iscyclic of prime order it is certainly solvable. ��Lemma A.1.4. (1) Let G be a solvable group and let H be a subgroup of G.Then H is a solvable group.

(2) Let G be a solvable group and let N be a normal subgroup of G. ThenG/N is a solvable group.

(3) Let G be a group and let N be a normal subgroup of G. If N and G/Nare solvable, then G is solvable.

Proof. Consider G = G0 ⊇ G1 ⊇ · · · ⊇ Gk ⊆ {1}. Then H = H0 ⊇ H1 ⊇· · · ⊇ Hk ⊇ {1} where Hi = H ∩ Gi . It is easy to check that Hi is a normalsubgroup of Hi−1, and then by standard theorems of group theory

Hi−1/Hi = (H ∩ Gi−1)/H ∩ Gi = (H ∩ Gi−1)/(H ∩ Gi−1) ∩ Gi

∼= (H ∩ Gi−1)Gi/Gi ⊆ Gi−1/Gi

is isomorphic to a subgroup of an abelian group so it is abelian.(2) Let π : G → G/N = Q be the quotient map and consider Q = Q0 ⊇

· · · ⊇ Qk = {1} where Qi = π(Gi ). Then Qi = Gi N/N ∼= Gi/Gi ∩ N , so

π(Gi−1)/π(Gi ) = (Gi−1 N/N )/(Gi N/N )

∼= (Gi−1/Gi−1 ∩ N )/(Gi/Gi ∩ N )

is isomorphic to a quotient of the abelian group Gi−1/Gi so it is abelian.(3) Let π : G → G/N = Q. Let N = N0 ⊇ · · · ⊇ N j = {1} and

Q = Q0 ⊇ · · · ⊇ Qk = {1} be as in Definition A.1.1. Then the sequence ofsubgroups

G = π−1(Q0) ⊇ π−1(Q1) ⊇ · · · ⊇ π−1(Qk) = N0 ⊇ N1 ⊇ · · · ⊇ N j = {1}shows that G is solvable. ��

A.1 Solvable Groups 197

So far, we have considered abstract solvable groups. However, when weconsider solvable subgroups of symmetric groups, we can get very precisedescriptions.

Lemma A.1.5. Let G be a subgroup of Sp of order divisible by p, p a prime.Then any nontrivial normal subgroup K of G has order divisible by p.

Proof. Since G has order divisible by p, it must contain an element of orderp, which must be a p-cycle.

Regard Sp as operating on T = {1, . . . , p}. Write T = T1 ∪ · · · ∪ Tq , adecomposition onto orbits of K . In other words, i, j, ∈ Tm for some m if andonly if there is some τ ∈ K with τ(i) = j . Now for any i, j ∈ T there is aσ ∈ G with σ(i) = j . This implies that the action of G permutes {T1, . . . , Tq}so in particular each Tm has the same cardinality. This cardinality then dividesp, so must be 1 or p, and cannot be 1 as K is nontrivial. Thus we see that Koperates transitively on T , so has order divisible by p (as the subgroup of Kfixing 1, say, has index p). ��Lemma A.1.6. Let G be a solvable subgroup of Sp of order divisible by p, pa prime. Then G contains a unique subgroup N of order p.

Proof. Let G = G0 ⊃ G1 ⊃ · · · ⊃ Gk = {1} as in Definition A.1.1. We provethe lemma by induction on k.

If k = 1, G = G0 ⊃ G1 = {1} and G is solvable, so G is abelian, andhence has a unique p-Sylow subgroup N , of order p.

Suppose the lemma is true for k − 1. Then G1 is a normal subgroup of G,so applying Lemma A.1.5 with K = G1 we see that G1 has order divisible byp, and then by the inductive hypothesis that G1 contains a unique subgroup Nof order p. Now let N ′ be any subgroup of G of order p. Then N ′ and N areboth p-Sylow subgroups, so are conjugate. Let N ′ = σ Nσ−1. As N ⊆ G1,N ′ = σ Nσ−1 ⊆ σ G1σ

−1 = G1 as G1 is normal in G, so N ′ = N , asrequired. ��

Let a group G operate on a set T . This action is effective if the only elementσ or G with σ(t) = t for all t ∈ T is σ = id.

Let a group G1 operate on a set T1, and a group G2 operate on a set T2.These operations are permutation isomorphic if there is an isomorphism ofgroups ψ : G1 → G2 and an isomorphism of sets (i.e., a bijection) � : T1 →T2 with �(σ(t)) = ψ(σ)(�(t)) for every σ ∈ G1, t ∈ T1.

Proposition A.1.7. Let G be a solvable group operating effectively and tran-sitively on a set T of cardinality p, p a prime. Then there is a subgroup H ofF∗

p such that this operation is permutation isomorphic to


G H ={ [

h n0 1

]| h ∈ H, n ∈ Fp

}

operating on

T ={ [

i1

]| i ∈ Fp

}

by left multiplication.

Proof. As we have observed, in this situation G has order divisible by p. (Lett ∈ T be arbitrary and consider {σ ∈ G | σ(t) = t}. Since G acts transitivelyon T , this subgroup of G has index p.)

Since G operates effectively on T , G is isomorphic to a subgroup of thesymmetric group Sp, so, by Lemma A.1.6, G contains a unique subgroup Nof order p, which is generated by a p-cycle. Reordering, if necessary, we have

that N and T are permutation isomorphic to{ [

1 n0 1

]| n ∈ Fp

}operating on

{ [i1

]| i ∈ Fp

}by left multiplication, as a suitable generator ν of N takes[

i1

]to

[i + 1

1

]mod p. For simplicity of notation, we simply identify N with

this group and T with this set.Since N is the unique subgroup of order p of G, it is certainly a normal

subgroup of G. Let σ ∈ G be arbitrary. Then σνσ−1 = νh for some h �= 0.

Let σ( [

01

] )=

[n1

]. Then, for any i ,

[n + ih

1

]= νih

( [n1

] )= σνiσ−1

( [n1

] )= σνi

( [01

] )= σ

( [i1

] ),

so

σ( [

i1

] )=

[h n0 1

] [i1

]

and hence

G ⊆{ [

h n0 1

]| h ∈ F∗

p, n ∈ Fp

}.

Thus, if H is the subgroup of F∗p given by

A.2 p-Groups 199

H ={

h ∈ F∗p |

[h 00 1

]∈ G

},

then G = G H as claimed. ��Corollary A.1.8. Let G be a subgroup of Sp, p a prime, of order divisible byp. The following are equivalent:

(1) G is solvable.(2) The order of G divides p(p − 1).(3) The order of G is at most p(p − 1).

Proof. (1) implies (2): If G is solvable, then, by Proposition A.1.7, G = G H

for some H and then |G| divides p(p − 1).(2) implies (3): Trivial.(3) implies (1): G has a p-Sylow subgroup N , and N is unique, as if G

had another p-Sylow subgroup N ′, then G would have order at least p2. Thenthe last part of the proof of Proposition A.1.7 shows that G = G H for someH . But G H ⊃ N ⊃ {1} with N normal in G. N is abelian and G H/N isisomorphic to H which is abelian, so G is solvable. ��

A.2 p-Groups

Throughout this section, p denotes a prime.

Definition A.2.1. A group G is a p-group if |G| is a power of p. �The center Z(G) of a group G is defined by Z(G) = {σ ∈ G | στ = τσ

for every τ ∈ G}.Lemma A.2.2. Let G be a p-group. Then G has a nontrivial center (i.e.,Z(G) �= {1}).Proof. Write G = C1 ∪ · · · ∪ Ck , a union of conjugacy classes. If σi ∈ Ci ,then |Ci | = [G : Hi ] where Hi = {τ ∈ G | τσiτ

−1 = σi }. If σi ∈ Z(G),then Ci = {σi } and |Ci | = 1 (and Hi = G). If σi /∈ Z(G), then Hi is a propersubgroup of G, so |Ci | = [Gi : Hi ] is a positive power of p. Of course |G| is apositive power of p. But one of the conjugacy classes, say C1, consists of {id}alone, and |C1| = 1 is not divisible by p. Now |G| = |C1| + · · ·+ |Ck |. Hencethere must exist other classes Ci with |Ci | not divisible by p, and hence with|Ci | = 1, and if Ci = {σi } then σi ∈ Z(G). ��


Corollary A.2.3. Let G be a p-group, |G| = pn. Then there is a sequence ofsubgroups G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = {1} with Gi a normal subgroup of Gof index pi , for each i = 1, . . . , n.

Proof. By induction on n. If n = 1, the result is trivial.Now suppose the result is true for all groups of order pn−1, and let G

be an arbitrary group of order pn . By Lemma A.2.2, the center Z(G) of Gis nontrivial. Since |Z(G)| divides |G|, |Z(G)| is also a power of p, so, inparticular, Z(G) contains an element of order p. The cyclic subgroup H of Ggenerated by that element is a subgroup of G of order p, and, since H ⊆ Z(G),H is a normal subgroup of G. Let Q = G/H be the quotient group, and letπ : G → Q be the canonical projection.

Now Q is a group of order pn−1, so by the inductive hypothesis there is asequence of subgroups Q = Q0 ⊃ Q1 ⊃ · · · ⊃ Qn−1 = {1} with Qi a normalsubgroup of Q of index pi for each i = 1, . . . , n − 1. Let Gi = π−1(Qi ) fori = 1, . . . , n − 1 and Gn = {1}. Then G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = {1}form a sequence as claimed. (Clearly Gi has index pi . Also, Gi is a normalsubgroup of G as if g0 ∈ Gi and g ∈ G, then gg0g−1 ∈ Gi as π(gg0g−1) =π(g)π(g0)(π(g))−1 ∈ Qi as Qi is a normal subgroup of Q.) ��Lemma A.2.4. Let G be a p-group. Then G is solvable.

Proof. By Corollary A.2.3, there is a sequence of subgroups G = G0 ⊃ G1 ⊃· · · ⊃ Gn = {1} with Gi a normal subgroup of G, and hence certainly a normalsubgroup of Gi−1, for each i . Furthermore, since [G : Gi ] = pi for each i ,[Gi−1 : Gi ] = p and hence |Gi−1/Gi | = p, and so Gi−1/Gi is cyclic of orderp and, in particular, is abelian, for each i . Hence, by Definition A.1.1, G issolvable. ��

A.3 Symmetric and Alternating Groups

In this section we prove several results about symmetric and alternatinggroups. We let Sn be the symmetric group on {1, . . . , n} and we let An ⊆ Sn

be the alternating group. We regard the elements of Sn as functions on thisset and recall that functions are composed by applying the rightmost functionfirst. Thus, for example, (1 2)(1 3) = (1 3 2). We recall that every elementof Sn can be written as a product of disjoint cycles, and that disjoint cyclescommute.

The first two results we prove allow us to conclude that certain subgroupsof Sn are in fact equal to Sn .

A.3 Symmetric and Alternating Groups 201

Lemma A.3.1. Let p be a prime and let G be a transitive subgroup of Sp thatcontains a transposition. Then G = Sp.

Proof. By renumbering if necessary, we may assume the transposition is τ =(1 2).

Since G acts transitively on {1, . . . , p}, it has order divisible by p andhence it has an element σ0 of order p. Since p is a prime, σ0 is a p-cycle.Thus there is some power σ = σ k

0 of σ0 with σ(1) = 2. By renumberingif necessary, we may then assume σ = (1 2 · · · p). Now direct calculationshows that

σ jτσ− j = (1 2 · · · p) j (1 2)(1 2 · · · p)− j = (( j + 1) ( j + 2))

for j = 0, . . . , p − 2. Direct calculation then shows that

(2 3)(1 2)(2 3) = (1 3),

(3 4)(1 3)(3 4) = (1 4),

...

((p − 1) p)(1 (p − 1))((p − 1) p) = (1 p)

and furthermore that

(1 k)(1 j)(1 k) = ( j k) for any j �= k.

Thus G contains every transposition. But Sp is generated by transpositions,so G = Sp. ��Lemma A.3.2. Let G be a transitive subgroup of Sn that contains an (n − 1)-cycle and a transposition. Then G = Sn.

Proof. Let ρ be the n −1-cycle. By renumbering if necessary, we may assumethat ρ = (1 2 · · · (n − 1)). Then the transposition is τ = (i j) for some i andj . Since G operates transitively on {1, . . . , n}, there is an element γ of G withγ ( j) = n. Then γ (i) = k for some k. Direct calculation shows that

γ τγ −1 = (k n) = τk and then that ρiτkρ−i = ((k + i) n) = τk+i ,

where k + i is taken mod n − 1. Finally, direct calculation shows that

τiτ jτ−1i = (i j) for any i �= j.

Thus G contains every transposition. But Sn is generated by transpositions,so G = Sn . ��


Next we show some results on the structure of An and Sn . We first need thefollowing technical lemma.

Lemma A.3.3. If G is a normal subgroup of An that contains a 3-cycle, thenG = An.

Proof. We shall first show that if G contains a single 3-cycle, then it containsevery 3-cycle. Note that if G contains a 3-cycle (i j k), it contains (i j k)−1 =(k j i).

By renumbering if necessary, we may assume that G contains the 3-cycle(1 2 3). Then for each i > 3 (observing that an element of order 2 is its owninverse) G also contains the elements

((1 2)(3 i))(3 2 1)((1 2)(3 i)) = (1 2 i),

((1 3)(2 i))(1 2 3)((1 3)(2 i)) = (1 3 i),

((2 3)(1 i))(3 2 1)((2 3)(1 i)) = (2 3 i);then for each distinct i, j > 3, G also contains the elements

(1 2 j)(2 i 1) = (1 i j),

(2 3 j)(3 i 2) = (2 i j),

(3 1 j)(1 i 3) = (3 i j);finally, for each distinct i, j, k > 3, G also contains the elements

(k i 1)(1 i j) = (i j k),

so G contains all 3-cycles, as claimed.But we now claim that any element of An can be written as a product of 3-

cycles. Since any element of An can be written as a product of an even numberof transpositions, to show this it suffices to show that the product of any twotranspositions can be written as a product of 3-cycles, and we see this from

(i j)(i k) = (i k j),

(i j)(k �) = (i k j)(i k �),

completing the proof. ��Theorem A.3.4. For n ≥ 5, An is a simple group (i.e., it has no normal sub-groups other than An and {1}).

A.3 Symmetric and Alternating Groups 203

Proof. Let n ≥ 5 and let G �= {1} be a normal subgroup of An . We will showG = An .

By Lemma A.3.3, it suffices to show that G contains a 3-cycle, so that iswhat we shall do.

Let σ ∈ G, σ �= 1. Then σ has order m > 1. Let p be a prime dividing mand let γ = σ m/p. Then γ has order p. Now γ can be written as a product ofdisjoint cycles, γ = γ1 · · · γk , and since γ has order p, each of γ1, . . . , γk is ap-cycle. Now we must consider several cases:

Case I: p ≥ 5: By renumbering if necessary, we may assume thatγ1 = (1 2 3 4 5 6 7 · · · p). Let τ1 = (1 2)(3 4). Then β1 = τ1γ1τ

−11 =

(1 4 3 5 6 7 · · · p 2) and γ1β−11 = (1 3 5 4 2). Letting β = τ1γ τ−1

1 andusing the fact that disjoint cycles commute, it easily follows that γβ−1 =(1 3 5 4 2). Let τ2 = (1 3)(2 4). Then τ2(1 3 5 4 2)τ−1

2 = (1 5 2 4 3) and(1 3 5 4 2)(1 5 2 4 3) = (1 4 5).

Case II: p = 3: If k = 1, i.e., there is only a single 3-cycle, there isnothing to do, so assume k ≥ 2. By renumbering if necessary, we may assumethat γ1 = (1 2 3) and γ2 = (4 5 6). Let γ12 = γ1γ2 = (1 2 3)(4 5 6). Let τ1 =(1 2)(3 4). Then δ12 = τ1γ12τ

−11 = (1 4 2)(3 5 6) and γ12δ

−112 = (1 3 4 2 5).

Letting δ = τ1γ τ−11 , it then follows, as in Case I, that γ δ−1 = (1 3 4 2 5).

Thus G contains a 5-cycle and we are back in Case I.

Case III: p = 2: Since we are in An , k is even, so k ≥ 2. By renumberingif necessary, we may assume that γ1 = (1 2) and γ2 = (3 4). Let γ12 =γ1γ2 = (1 2)(3 4). Let ρ = (1 2 3). Then ζ12 = ργ12ρ

−1 = (1 4)(2 3) andγ12ζ

−112 = (1 3)(2 4). Letting ζ = ργρ−1, it then follows, as in Case I, that

γ ζ−1 = θ = (1 3)(2 4). Let τ3 = (1 3)(2 5). (Note here is where we needn ≥ 5.) Then τ3θτ−1

3 = θ ′ = (1 3)(4 5). But then θθ ′ = (2 4 5). ��Theorem A.3.5. For n ≥ 5, the only normal subgroups of Sn are Sn, An,and {1}.Proof. Let n ≥ 5 and let G �= {1} be a normal subgroup of Sn . We willshow G ⊇ An . Since An is a subgroup of Sn of index 2, this shows that G =An or Sn .

The proof of this is almost identical to the proof of Theorem A.3.4. Webegin in the same way. Let σ ∈ G, σ �= 1. Then σ has order m > 1. Let p bea prime dividing m and let γ = σ m/p. Then γ has order p.

If we are in Case I, Case II, or Case III with k ≥ 2 of the proof of TheoremA.3.4, we conclude in exactly the same way that G ⊇ An . This leaves only thecase where p = 2 and k = 1, i.e., where γ is a single transposition. But alltranspositions in Sn are conjugate, and Sn is generated by transpositions, so inthis case G = Sn . ��


Corollary A.3.6. For n ≥ 5, neither An nor Sn is solvable.

Proof. Let n ≥ 5. By the definition of a solvable group (Definition A.1.1), ifG = An or Sn were solvable, it would have a composition series with abelianquotients. But by Theorem A.3.4, the only possible composition series for An

is An ⊃ {1}, and by Theorem A.3.5, the only possible composition series forSn are Sn ⊃ {1} and Sn ⊃ An ⊃ {1}, and An is not abelian. ��

B

A Lemma on Constructing Fields

In Section 2.2 we showed: If f (X) ∈ F[X ] is an irreducible polynomial, thenF[X ]/〈 f (X)〉 is a field. We did so by using arguments particular to polynomi-als. In this appendix we shall show that this result follow from more generalresults in ring theory. In particular, in this section we show how to constructfields from integral domains.

Definition B.1.1. Let R be an integral domain.(1) An ideal I ⊂ R is a maximal ideal if I ⊆ J ⊆ R, J an ideal, implies

J = I or J = R.(2) An ideal I ⊂ R is a prime ideal if ab ∈ I , with a, b ∈ R, implies a ∈ I

or b ∈ I . �Lemma B.1.2. (1) Let R be an integral domain. Then every maximal ideal isprime.

(2) Let R be a PID (principal ideal domain). Then every prime ideal ismaximal.

Proof. (1) Let I be a maximal ideal and let ab = i ∈ I . If a ∈ I , thereis nothing to prove. Otherwise, let J = 〈a, I 〉 (the ideal generated by a andI ). Then I ⊂ J and I is maximal, so J = R. In particular, 1 ∈ J . Thenxa + i ′ = 1 for some x ∈ R and i ′ ∈ I . Then

b = 1 · b = (xa + i ′)b = X (ab) + i ′b = i X + i ′b ∈ I.

(2) Suppose I is a prime ideal and J is an ideal with I ⊆ J . Then I = 〈p〉for some p ∈ R and J = 〈q〉 for some q ∈ R. Since I ⊆ J , p ∈ J so p = qrfor some r ∈ R. Since I is prime, we conclude that:

(a) q ∈ I in which case J = 〈q〉 ⊆ I so J = I ; or

206 B A Lemma on Constructing Fields

(b) r ∈ I so r = pr ′ for some r ′ ∈ R in which case p = qr = qpr ′ =p(qr ′) so 1 = qr ′ and hence 1 ∈ J , so J = R. ��Lemma B.1.3. Let R be an integral domain and I ⊂ R an ideal. Then I ismaximal if and only if R/I is a field.

Proof. Let π : R → R/I be the canonical projection. First suppose I ismaximal. Let a ∈ R/I , a �= 0. Then a = π(a) for some a ∈ R, a /∈ I . SinceI is maximal, R = 〈a, I 〉, so 1 = xa + i for some x ∈ R and some i ∈ I .Then, if x = π(x),

1 = π(1) = π(xa + i) = x a + 0 = x a,

so a is invertible.Now suppose I is not maximal, and let I ⊂ J ⊂ R. Let j ∈ J , j /∈ I , and

let j = π( j). We claim j is not invertible. For suppose j k = 1, j, k ∈ R/I ,and let k = π(k). Then jk = 1 + i for some i ∈ I , so 1 = jk − i and hence1 ∈ J , a contradiction. ��Corollary B.1.4. Let R be a PID and let I = 〈i〉 be the ideal generated by theelement i of R. Then R/I is a field if and only if i is irreducible.

Proof. In a PID, an element i is irreducible if and only if i is prime, and anideal I = 〈i〉 is a prime ideal if and only if the element i is prime, so thecorollary follows directly from Lemma B.1.2 and Lemma B.1.3. ��Corollary B.1.5. Let f (X) ∈ F[X ] be an irreducible polynomial. ThenF[X ]/〈 f (X)〉 is a field.

Proof. Since F[X ] is a PID, this is a special case of Corollary B.1.4. ��

C

A Lemma from Elementary Number Theory

In this appendix, we prove a lemma that is used in the proof of Theorem 4.7.1.

Lemma C.1.1. Let p be a prime and let t be a positive integer. Then there areinfinitely many primes congruent to 1(mod pt ).

Proof. We begin by recalling the following definition and facts: Let q bea prime and let a be an integer relatively prime to q. Then ordq(a), theorder of a(mod q), is defined to be the smallest positive integer x such thatax ≡ 1(mod q). Then, for an arbitrary integer y, ay ≡ 1(mod q) if and only ifordq(a) divides y. Also, since, by Fermat’s Little Theorem, aq−1 ≡ 1(mod q),we have that ordq(a) divides q − 1.

We now proceed with the proof, which we divide into two cases: p = 2and p odd.

Case I: p = 2. For a nonnegative integer k, let Fk = 22k + 1. Note that, ifj < k, Fj = 22 j +1 divides 22k −1 = (22 j

)2k− j −1 = Fk −2, so gcd(Fj , Fk) =gcd(Fj , 2) = 1, i.e., {F1, F2, . . . } are pairwise relatively prime. Thus, if qk isa prime factor of Fk , then {q1, q2, . . . } are all distinct.

Now let q = qs for s ≥ t − 1. By definition, q divides Fs , i.e., 22s +1 ≡ 0(mod q), 22s ≡ −1(mod q), and hence 22s+1 ≡ 1(mod q). Thus ordq(2)

divides 2s+1 but does not divide 2s , so ordq(2) = 2s+1. Then, as we haveobserved above, 2s+1 divides q − 1, so q ≡ 1(mod 2s+1) and hence q ≡1(mod 2t ).

Thus we see that {qt−1, qt , . . . } are distinct primes congruent to 1(mod 2t ).

Case II: p odd. In this case we proceed by contradiction. Assume there areonly finitely many primes ≡ 1(mod pt ), and let these be q1, . . . , qr . Let

a = 2q1 · · · qr , c = a pt−1, N = a pt − 1 = cp − 1,

208 C A Lemma from Elementary Number Theory

and write N = (c − 1)M where M = cp−1 + · · · + 1.

Claim: c − 1 and M are relatively prime.

Proof of claim: M = cp−1 + · · · + 1 = (cp−1 − 1) + (cp−2 − 1) + · · · +(c − 1) +p, and each term except possibly the last one is divisible by c − 1, sogcd(c − 1, M) = gcd(c − 1, p) = 1 or p. But a ≡ 2(mod p) so c ≡2pt−1

(mod p). By Fermat’s Little Theorem, 2p−1 ≡ 1(mod p), and pt−1 ≡1(mod p − 1), so c ≡ 21 ≡ 2(mod p) and so c − 1 ≡ 1(mod p). Thus p doesnot divide c − 1 and so gcd(c − 1, M) = gcd(c − 1, p) = 1.

Now let q be any prime dividing M . Then q divides N = cp − 1,so cp ≡ 1(mod q), i.e., a pt ≡ 1(mod q). So ordq(a) divides pt , i.e.,ordq(a) = 1, p, . . . , pt−1 or pt . But if ordq(a) = 1, p, . . . , or pt−1, thena pt−1 ≡ 1(mod q), i.e., c ≡ 1(mod q), so q divides c − 1, contradicting thefact that c − 1 and M are relatively prime. Hence ordq(a) = pt . Then, as wehave observed above, pt divides q − 1, i.e., q ≡ 1(mod pt ).

But certainly q does not divides a, while each of q1, . . . , qr does divide a,so q �= q1, . . . , qr , contradicting the hypothesis that q1, . . . , qr are all theprimes congruent to 1(mod pt ). Hence there must be infinitely many suchprimes. ��

Index

17-gon 105

Abel 99adjoining 17algebraic

integer 42number 18, 20number field 20

algebraicallyclosed 159independent 174

algorithmfor computing Galois groups 128algorithm for factoring polynomials 92

Artin 30Artin–Schreier extension 83

basisnormal, 49

character 29characteristic 9closure

algebraic, 155, 158Galois, 152normal, 152separable, 147, 149

conjugateGalois, 27

curveFermat, 186, 191plane, 185rational, 186

degree 9, 16, 17, 19, 23, 27, 60, 61,66, 117, 118, 148

inseparable, 149separable, 149

Dirichlet 29, 123discriminant 129division algorithm 11domain

Euclidean, 11integral, 10, 203principal ideal, 11, 203

duplicating the cube 103

Eisenstein’s Criterion 91element

algebraic, 18, 21inseparable, 25irreducible, 204primitive, 47, 66, 68purely inseparable, 147, 148separable, 25, 145transcendental, 173

extension 9abelian, 73, 122, 123algebraic, 18, 20, 21

210 Index

extension (continued)completely transcendental, 173composite, 170finite, 9Galois, 27, 28, 32, 35, 60, 61, 63, 161,

162, 170, 171inseparable, 25, 55nonsimple, 69normal, 25, 26, 28, 32, 149, 151, 152,

161purely inseparable, 147–149purely transcendental, 173quadratic, 108radical, 97, 112separable, 25, 26, 28, 54, 55, 143, 145,

146, 161simple, 47, 66, 67symmetric, 122, 123transcendental, 173

extension by radicals see extension,radical

extensionsconjugate, 34disjoint, 60–62, 96, 170linearly disjoint, 66

Fermat prime 105Fermat’s Little Theorem 10field 7, 13, 204

algebraically closed, 155, 158composite, 16cyclotomic, 93, 108, 112, 114, 123extension, 14, 15finite, 55, 56fixed, 64function, 185Kummer, 76, 112perfect, 26, 55quadratic, 111, 112splitting, 22, 24–26, 28, 151, 161,

162Frobenius map 10, 55, 56, 143FTGT see Fundamental Theorem of

Galois Theory

functionsymmetric, 45

Fundamental Theorem of Algebra 159Fundamental Theorem of Galois Theory

2, 27, 32, 37, 57, 164Fundamental Theorem of Infinite Galois

Theory 164

Galois 99Gauss 105, 112Gauss’s Lemma 90gcd see greatest common divisorgreatest common divisor 11, 15group

abelian, 122alternating, 198, 200cyclic, 12Galois, 1, 26, 32, 36, 37, 45, 58, 60,

63, 64, 73–75, 77, 96, 112, 113,115, 117, 120, 122, 123, 126–128,168, 171

nonsolvable, 202p-group, 197, 198permutation, 35simple, 200solvable, 98, 193–195, 197, 198symmetric, 123, 127, 195, 197–199,

201topological, 163

Hilbert’s Theorem 90 81

idealmaximal, 203, 204prime, 203

inseparabilitydegree of, 147, 150

invariance of gcd under field extensions15

Kronecker 14, 92Kronecker–Weber Theorem 123

Lagrange Interpolation Formula 92Landau 95Lindemann 103

Index 211

Luroth 183

Mobius Inversion Formula 60, 97

n-powerless 74–76, 114Newton’s Identities 51norm 79, 80normal basis 69, 70, 72

polynomialcyclotomic, 91, 93, 94elementary symmetric, 46inseparable, 25, 55irreducible, 58, 90–92, 94, 116,

204minimum, 18, 19, 27, 65, 80,

151primitive, 90radical, 97, 112separable, 25, 26, 28, 54, 145,

146, 161, 162symmetric, 45

polynomial ring 11Pythagorean triple 191

solvable by radicals 97–99not, 99, 100

splits 22squaring the circle 103straightedge and compass 101, 103,

105

Theorem on Natural Irrationalities 62,158

topologycompact-open, 170Krull, 163, 167, 168, 170product, 170

trace 79, 80transcendence

basis 175degree 177

transcendental extensionsimple, 181

transformation, linear 19trisecting the angle 103

van der Waerden 123

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